Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's relates the coefficients of a polynomial to sums and products of its roots. But wait, the equation here isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So maybe I need to manipulate the equation to convert it into a polynomial equation first?

Let me write down the original equation again:

1/x + 5x² = 6x - 24.

Okay, so to eliminate the fraction, I can multiply both sides of the equation by x. That should get rid of the denominator. Let me try that:

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

So bringing everything to the left side:

1 + 5x³ - 6x² + 24x = 0

Which can be written as:

5x³ - 6x² + 24x + 1 = 0

Wait, hold on. Let me check the signs again. If I subtract the right side from both sides:

Left side - Right side = 0:

(1 + 5x³) - (6x² - 24x) = 0

Which is 1 + 5x³ - 6x² + 24x = 0. Yes, that's correct. So rearranged:

5x³ - 6x² + 24x + 1 = 0

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0.

But wait, the original equation had 1/x, so when we multiplied by x, we assumed that x ≠ 0. Therefore, we need to check if x = 0 is a solution. Plugging x = 0 into the original equation: 1/0 is undefined, so x = 0 is not a solution. Therefore, our transformed equation 5x³ - 6x² + 24x + 1 = 0 should have all the roots of the original equation.

But the problem states that a, b, c are the roots of the original equation. However, the transformed equation is a cubic, so it has three roots. Therefore, a, b, c are the roots of this cubic equation. Therefore, we can apply Vieta's formula here.

Vieta's formula for a cubic equation ax³ + bx² + cx + d = 0 states that the sum of the roots is -b/a.

So in our case, the cubic equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficients are:

a_cubic = 5 (coefficient of x³)

b_cubic = -6 (coefficient of x²)

c_cubic = 24 (coefficient of x)

d_cubic = 1 (constant term)

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5.

Wait, so is that the answer? 6/5? But let me make sure I didn't make any mistakes in transforming the original equation.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0. Yes, that seems correct.

Therefore, the cubic equation is correct, and Vieta's formula gives the sum of roots as 6/5. Therefore, a + b + c = 6/5.

But wait, hold on. Let me check again. Maybe I made a mistake in the sign when applying Vieta's formula. Let me recall Vieta's formula for cubic equations.

For a general cubic equation: ax³ + bx² + cx + d = 0, the sum of roots is -b/a.

In our case, the equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficient of x² is -6, so the sum of roots is -(-6)/5 = 6/5. Yes, that's correct.

But let me think again. Suppose I have a cubic equation with roots a, b, c. Then:

(x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc.

But our equation is 5x³ - 6x² + 24x + 1 = 0. To match the standard form, we can divide both sides by 5:

x³ - (6/5)x² + (24/5)x + 1/5 = 0.

Comparing with the standard cubic equation x³ - (a + b + c)x² + (ab + ac + bc)x - abc = 0, we can see that:

-(a + b + c) = -6/5 => a + b + c = 6/5.

Yes, that's correct. Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "Compute a + b + c, given that a, b, and c are the roots of [equation]." So according to this, the answer is 6/5.

But let me check if there's any possibility that the original equation might have more roots or something? For example, when we multiplied both sides by x, we might have introduced extraneous roots? Wait, but we already checked that x = 0 is not a solution, so the transformed equation should have exactly the same roots as the original equation.

But let me check with an example. Suppose there is a root x = k of the original equation. Then, multiplying by x gives 1 + 5k³ = 6k² - 24k, which is equivalent to 5k³ - 6k² + 24k + 1 = 0. Therefore, k is a root of the cubic equation. Conversely, if x is a root of the cubic equation, then 5x³ - 6x² + 24x + 1 = 0, which implies 1 + 5x³ = 6x² - 24x, so dividing both sides by x (since x ≠ 0) gives 1/x + 5x² = 6x - 24, which is the original equation. Therefore, the roots of the cubic equation are exactly the roots of the original equation. Therefore, there are no extraneous roots introduced.

Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "a, b, and c are the roots". However, a cubic equation can have up to three roots (real or complex). But the original equation is 1/x + 5x² = 6x - 24. Depending on the context, maybe the problem assumes that a, b, c are real roots? But even so, Vieta's formula applies regardless of whether the roots are real or complex.

But maybe I need to check if the cubic equation actually has three real roots? Let me think.

But the problem doesn't specify, so I think we can assume that a, b, c are all roots (real or complex) of the equation, and Vieta's formula still applies. Therefore, the answer is 6/5.

But let me check if I did the algebra correctly again. Let me re-derive the equation step by step.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0.

Yes, that's correct.

Alternatively, maybe I can check with substituting a value. Let me try to see if there are any obvious roots. For example, let me try x = -1:

Left side of original equation: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 + 5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. Not equal.

x = 1: Left side: 1/1 + 5(1)^2 = 1 + 5 = 6. Right side: 6(1) -24 = 6 -24 = -18. Not equal.

x = 2: Left side: 1/2 + 5*(4) = 0.5 + 20 = 20.5. Right side: 6*2 -24 = 12 -24 = -12. Not equal.

x = -2: Left side: 1/(-2) + 5*(4) = -0.5 + 20 = 19.5. Right side: 6*(-2) -24 = -12 -24 = -36. Not equal.

x = 1/2: Left side: 1/(1/2) + 5*(1/2)^2 = 2 + 5*(1/4) = 2 + 1.25 = 3.25. Right side: 6*(1/2) -24 = 3 -24 = -21. Not equal.

Hmm, maybe there are no rational roots? Let me check using Rational Root Theorem on the cubic equation 5x³ - 6x² + 24x + 1 = 0.

Possible rational roots are ±1, ±1/5. Let me test x = -1/5:

5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = (-1/25) - 6/25 - 24/5 +1 = (-7/25) - 24/5 +1. Converting to 25 denominator: -7/25 - 120/25 + 25/25 = (-7 -120 +25)/25 = (-102)/25 ≠ 0.

x = 1/5:

5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 - 6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = -1/5 + 24/5 +1 = (23/5) +1 = 28/5 ≠ 0.

x = 1: 5 -6 +24 +1 = 24 ≠ 0.

x = -1: -5 -6 -24 +1 = -34 ≠ 0.

So no rational roots. Therefore, maybe the cubic doesn't factor nicely, but that's okay. Vieta's formula still applies regardless.

Therefore, unless there's some mistake in my reasoning, the answer should be 6/5.

But let me check once more. Maybe I made a mistake in the sign when moving terms. Let me go back:

Original equation after multiplying by x: 1 + 5x³ = 6x² -24x.

So bringing all terms to the left: 1 + 5x³ -6x² +24x = 0. Which is 5x³ -6x² +24x +1 = 0. Yes, that's correct.

Alternatively, if I write it as 5x³ -6x² +24x +1 = 0.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5.

Therefore, I think that is correct.

But maybe the problem is expecting integer or fractional answer? 6/5 is 1.2, which is a fraction.

Alternatively, maybe I made a mistake in the original equation transformation? Let me check again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then moving all terms to left: 5x³ -6x² +24x +1 = 0.

Yes.

Alternatively, maybe the problem is written differently? Let me check the original problem again:

"Compute a+b+c, given that a, b, and c are the roots of 1/x + 5x² = 6x - 24."

Yes, that's what I used.

Alternatively, maybe there's a typo in my calculation? Let me check once again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then subtract 6x² and add 24x to both sides: 1 + 5x³ -6x² +24x = 0.

Yes. So 5x³ -6x² +24x +1 = 0.

Therefore, the coefficients are correct.

Therefore, unless there's some other trick here, I think the answer is 6/5.

But let me think again. Maybe the original equation is supposed to be 1/x + 5x² = 6x - 24, but perhaps there's a different interpretation? For example, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, the original problem is written as 1/x + 5x² = 6x -24, so I think the correct interpretation is as written.

Alternatively, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, given the original problem is written as \frac{1}{x} + 5x^2 = 6x - 24, so it's definitely 1/x + 5x² equals that.

Therefore, I think my approach is correct.

Alternatively, maybe the problem is expecting me to consider that multiplying by x could introduce extraneous roots? But as discussed earlier, since x ≠ 0, and we checked that x=0 is not a solution, so all roots of the cubic equation are valid roots of the original equation.

Therefore, I think the answer is 6/5.

But just to be thorough, let me check if there's another way to approach this problem. For example, maybe using substitution or something else.

Alternatively, suppose I let y = x, then the equation is 1/y + 5y² = 6y -24. But that's the same as before.

Alternatively, maybe rearrange the original equation:

1/x = 6x -24 -5x².

Then, 1/x = -5x² +6x -24.

Then, cross multiplying: 1 = (-5x² +6x -24)x = -5x³ +6x² -24x.

Then, bringing all terms to left: 1 +5x³ -6x² +24x = 0, which is same as before: 5x³ -6x² +24x +1 = 0.

Same equation. Therefore, same result.

Therefore, I think my answer is correct.

Therefore, the sum a + b + c is 6/5, so the answer is \boxed{\dfrac{6}{5}}.

**Final Answer**
\boxed{\dfrac{6}{5}}

To solve the problem, we are given the equation:

$$
\frac{1}{x} + 5x^2 = 6x - 24
$$

and asked to compute $ a + b + c $, where $ a, b, c $ are the roots of this equation.

---

### Step 1: Eliminate the Fraction

The equation contains a term $ \frac{1}{x} $, which suggests that the equation is not a polynomial. To make it a polynomial, we multiply both sides of the equation by $ x $, assuming $ x \ne 0 $ (since $ x = 0 $ would make the original equation undefined):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into a Polynomial Equation

Bring all terms to one side to form a standard polynomial equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is a cubic polynomial in $ x $, and we are told that $ a, b, c $ are the roots of the original equation. Since we have transformed the original equation into a cubic polynomial, and verified that $ x = 0 $ is not a solution, the roots of the cubic equation are exactly the roots of the original equation.

---

### Step 3: Apply Vieta's Formula

For a cubic polynomial of the form:

$$
Ax^3 + Bx^2 + Cx + D = 0
$$

the sum of the roots $ a + b + c $ is given by:

$$
a + b + c = -\frac{B}{A}
$$

In our case, the polynomial is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

So, $ A = 5 $, $ B = -6 $, and thus:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{6}{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of a + b + c, where a, b, and c are the roots of the equation given: 1/x + 5x² = 6x - 24. Hmm, let me think about how to approach this.

First, I remember that when dealing with equations involving roots, Vieta's formulas are usually helpful. Vieta's relates the coefficients of a polynomial to sums and products of its roots. But wait, the equation here isn't a polynomial equation yet. It has a term with 1/x, which complicates things. So maybe I need to manipulate the equation to convert it into a polynomial equation first?

Let me write down the original equation again:

1/x + 5x² = 6x - 24.

Okay, so to eliminate the fraction, I can multiply both sides of the equation by x. That should get rid of the denominator. Let me try that:

Multiplying both sides by x:

x*(1/x) + x*(5x²) = x*(6x) - x*(24)

Simplifying each term:

1 + 5x³ = 6x² - 24x

Now, let me bring all terms to one side to set the equation to zero:

5x³ - 6x² + 24x + 1 = 0

Wait, let me check that again. Starting from:

Left side after multiplying by x: 1 + 5x³

Right side after multiplying by x: 6x² - 24x

So bringing everything to the left side:

1 + 5x³ - 6x² + 24x = 0

Which can be written as:

5x³ - 6x² + 24x + 1 = 0

Wait, hold on. Let me check the signs again. If I subtract the right side from both sides:

Left side - Right side = 0:

(1 + 5x³) - (6x² - 24x) = 0

Which is 1 + 5x³ - 6x² + 24x = 0. Yes, that's correct. So rearranged:

5x³ - 6x² + 24x + 1 = 0

So now we have a cubic equation: 5x³ - 6x² + 24x + 1 = 0.

But wait, the original equation had 1/x, so when we multiplied by x, we assumed that x ≠ 0. Therefore, we need to check if x = 0 is a solution. Plugging x = 0 into the original equation: 1/0 is undefined, so x = 0 is not a solution. Therefore, our transformed equation 5x³ - 6x² + 24x + 1 = 0 should have all the roots of the original equation.

But the problem states that a, b, c are the roots of the original equation. However, the transformed equation is a cubic, so it has three roots. Therefore, a, b, c are the roots of this cubic equation. Therefore, we can apply Vieta's formula here.

Vieta's formula for a cubic equation ax³ + bx² + cx + d = 0 states that the sum of the roots is -b/a.

So in our case, the cubic equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficients are:

a_cubic = 5 (coefficient of x³)

b_cubic = -6 (coefficient of x²)

c_cubic = 24 (coefficient of x)

d_cubic = 1 (constant term)

Therefore, the sum of the roots a + b + c = -b_cubic / a_cubic = -(-6)/5 = 6/5.

Wait, so is that the answer? 6/5? But let me make sure I didn't make any mistakes in transforming the original equation.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0. Yes, that seems correct.

Therefore, the cubic equation is correct, and Vieta's formula gives the sum of roots as 6/5. Therefore, a + b + c = 6/5.

But wait, hold on. Let me check again. Maybe I made a mistake in the sign when applying Vieta's formula. Let me recall Vieta's formula for cubic equations.

For a general cubic equation: ax³ + bx² + cx + d = 0, the sum of roots is -b/a.

In our case, the equation is 5x³ - 6x² + 24x + 1 = 0. Therefore, the coefficient of x² is -6, so the sum of roots is -(-6)/5 = 6/5. Yes, that's correct.

But let me think again. Suppose I have a cubic equation with roots a, b, c. Then:

(x - a)(x - b)(x - c) = x³ - (a + b + c)x² + (ab + ac + bc)x - abc.

But our equation is 5x³ - 6x² + 24x + 1 = 0. To match the standard form, we can divide both sides by 5:

x³ - (6/5)x² + (24/5)x + 1/5 = 0.

Comparing with the standard cubic equation x³ - (a + b + c)x² + (ab + ac + bc)x - abc = 0, we can see that:

-(a + b + c) = -6/5 => a + b + c = 6/5.

Yes, that's correct. Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "Compute a + b + c, given that a, b, and c are the roots of [equation]." So according to this, the answer is 6/5.

But let me check if there's any possibility that the original equation might have more roots or something? For example, when we multiplied both sides by x, we might have introduced extraneous roots? Wait, but we already checked that x = 0 is not a solution, so the transformed equation should have exactly the same roots as the original equation.

But let me check with an example. Suppose there is a root x = k of the original equation. Then, multiplying by x gives 1 + 5k³ = 6k² - 24k, which is equivalent to 5k³ - 6k² + 24k + 1 = 0. Therefore, k is a root of the cubic equation. Conversely, if x is a root of the cubic equation, then 5x³ - 6x² + 24x + 1 = 0, which implies 1 + 5x³ = 6x² - 24x, so dividing both sides by x (since x ≠ 0) gives 1/x + 5x² = 6x - 24, which is the original equation. Therefore, the roots of the cubic equation are exactly the roots of the original equation. Therefore, there are no extraneous roots introduced.

Therefore, the sum of the roots is indeed 6/5.

But wait, the problem says "a, b, and c are the roots". However, a cubic equation can have up to three roots (real or complex). But the original equation is 1/x + 5x² = 6x - 24. Depending on the context, maybe the problem assumes that a, b, c are real roots? But even so, Vieta's formula applies regardless of whether the roots are real or complex.

But maybe I need to check if the cubic equation actually has three real roots? Let me think.

But the problem doesn't specify, so I think we can assume that a, b, c are all roots (real or complex) of the equation, and Vieta's formula still applies. Therefore, the answer is 6/5.

But let me check if I did the algebra correctly again. Let me re-derive the equation step by step.

Original equation: 1/x + 5x² = 6x - 24.

Multiply both sides by x: 1 + 5x³ = 6x² - 24x.

Bring all terms to left: 5x³ - 6x² + 24x + 1 = 0.

Yes, that's correct.

Alternatively, maybe I can check with substituting a value. Let me try to see if there are any obvious roots. For example, let me try x = -1:

Left side of original equation: 1/(-1) + 5*(-1)^2 = -1 + 5(1) = -1 + 5 = 4.

Right side: 6*(-1) -24 = -6 -24 = -30. Not equal.

x = 1: Left side: 1/1 + 5(1)^2 = 1 + 5 = 6. Right side: 6(1) -24 = 6 -24 = -18. Not equal.

x = 2: Left side: 1/2 + 5*(4) = 0.5 + 20 = 20.5. Right side: 6*2 -24 = 12 -24 = -12. Not equal.

x = -2: Left side: 1/(-2) + 5*(4) = -0.5 + 20 = 19.5. Right side: 6*(-2) -24 = -12 -24 = -36. Not equal.

x = 1/2: Left side: 1/(1/2) + 5*(1/2)^2 = 2 + 5*(1/4) = 2 + 1.25 = 3.25. Right side: 6*(1/2) -24 = 3 -24 = -21. Not equal.

Hmm, maybe there are no rational roots? Let me check using Rational Root Theorem on the cubic equation 5x³ - 6x² + 24x + 1 = 0.

Possible rational roots are ±1, ±1/5. Let me test x = -1/5:

5*(-1/5)^3 -6*(-1/5)^2 +24*(-1/5) +1 = 5*(-1/125) -6*(1/25) -24/5 +1 = (-1/25) - 6/25 - 24/5 +1 = (-7/25) - 24/5 +1. Converting to 25 denominator: -7/25 - 120/25 + 25/25 = (-7 -120 +25)/25 = (-102)/25 ≠ 0.

x = 1/5:

5*(1/5)^3 -6*(1/5)^2 +24*(1/5) +1 = 5*(1/125) -6*(1/25) +24/5 +1 = 1/25 - 6/25 + 24/5 +1 = (-5/25) + 24/5 +1 = -1/5 + 24/5 +1 = (23/5) +1 = 28/5 ≠ 0.

x = 1: 5 -6 +24 +1 = 24 ≠ 0.

x = -1: -5 -6 -24 +1 = -34 ≠ 0.

So no rational roots. Therefore, maybe the cubic doesn't factor nicely, but that's okay. Vieta's formula still applies regardless.

Therefore, unless there's some mistake in my reasoning, the answer should be 6/5.

But let me check once more. Maybe I made a mistake in the sign when moving terms. Let me go back:

Original equation after multiplying by x: 1 + 5x³ = 6x² -24x.

So bringing all terms to the left: 1 + 5x³ -6x² +24x = 0. Which is 5x³ -6x² +24x +1 = 0. Yes, that's correct.

Alternatively, if I write it as 5x³ -6x² +24x +1 = 0.

So Vieta's formula says sum of roots is -(-6)/5 = 6/5.

Therefore, I think that is correct.

But maybe the problem is expecting integer or fractional answer? 6/5 is 1.2, which is a fraction.

Alternatively, maybe I made a mistake in the original equation transformation? Let me check again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then moving all terms to left: 5x³ -6x² +24x +1 = 0.

Yes.

Alternatively, maybe the problem is written differently? Let me check the original problem again:

"Compute a+b+c, given that a, b, and c are the roots of 1/x + 5x² = 6x - 24."

Yes, that's what I used.

Alternatively, maybe there's a typo in my calculation? Let me check once again:

Original equation: 1/x + 5x² = 6x -24.

Multiply both sides by x: 1 + 5x³ = 6x² -24x.

Yes. Then subtract 6x² and add 24x to both sides: 1 + 5x³ -6x² +24x = 0.

Yes. So 5x³ -6x² +24x +1 = 0.

Therefore, the coefficients are correct.

Therefore, unless there's some other trick here, I think the answer is 6/5.

But let me think again. Maybe the original equation is supposed to be 1/x + 5x² = 6x - 24, but perhaps there's a different interpretation? For example, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, the original problem is written as 1/x + 5x² = 6x -24, so I think the correct interpretation is as written.

Alternatively, maybe the equation is supposed to be 1/(x + 5x²) = 6x -24? But that would be a different equation. However, given the original problem is written as \frac{1}{x} + 5x^2 = 6x - 24, so it's definitely 1/x + 5x² equals that.

Therefore, I think my approach is correct.

Alternatively, maybe the problem is expecting me to consider that multiplying by x could introduce extraneous roots? But as discussed earlier, since x ≠ 0, and we checked that x=0 is not a solution, so all roots of the cubic equation are valid roots of the original equation.

Therefore, I think the answer is 6/5.

But just to be thorough, let me check if there's another way to approach this problem. For example, maybe using substitution or something else.

Alternatively, suppose I let y = x, then the equation is 1/y + 5y² = 6y -24. But that's the same as before.

Alternatively, maybe rearrange the original equation:

1/x = 6x -24 -5x².

Then, 1/x = -5x² +6x -24.

Then, cross multiplying: 1 = (-5x² +6x -24)x = -5x³ +6x² -24x.

Then, bringing all terms to left: 1 +5x³ -6x² +24x = 0, which is same as before: 5x³ -6x² +24x +1 = 0.

Same equation. Therefore, same result.

Therefore, I think my answer is correct.

Therefore, the sum a + b + c is 6/5, so the answer is \boxed{\dfrac{6}{5}}.

**Final Answer**
\boxed{\dfrac{6}{5}}

To solve the problem, we are given the equation:

$$
\frac{1}{x} + 5x^2 = 6x - 24
$$

and asked to compute $ a + b + c $, where $ a, b, c $ are the roots of this equation.

---

### Step 1: Eliminate the Fraction

The equation contains a term $ \frac{1}{x} $, which suggests that the equation is not a polynomial. To make it a polynomial, we multiply both sides of the equation by $ x $, assuming $ x \ne 0 $ (since $ x = 0 $ would make the original equation undefined):

$$
x \cdot \left( \frac{1}{x} + 5x^2 \right) = x \cdot (6x - 24)
$$

Simplifying both sides:

$$
1 + 5x^3 = 6x^2 - 24x
$$

---

### Step 2: Rearrange into a Polynomial Equation

Bring all terms to one side to form a standard polynomial equation:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

This is a cubic polynomial in $ x $, and we are told that $ a, b, c $ are the roots of the original equation. Since we have transformed the original equation into a cubic polynomial, and verified that $ x = 0 $ is not a solution, the roots of the cubic equation are exactly the roots of the original equation.

---

### Step 3: Apply Vieta's Formula

For a cubic polynomial of the form:

$$
Ax^3 + Bx^2 + Cx + D = 0
$$

the sum of the roots $ a + b + c $ is given by:

$$
a + b + c = -\frac{B}{A}
$$

In our case, the polynomial is:

$$
5x^3 - 6x^2 + 24x + 1 = 0
$$

So, $ A = 5 $, $ B = -6 $, and thus:

$$
a + b + c = -\frac{-6}{5} = \frac{6}{5}
$$

---

### Final Answer

$$
\boxed{\dfrac{6}{5}}
$$

Solution: We want to apply Vieta's formulas, but the given equation is not a polynomial equation, because of the $\frac1x$ term. To turn this equation into an equivalent polynomial equation, we multiply both sides by $x$ and rearrange: \[\begin{aligned} 1+5x^3 &= 6x^2 - 24x \\ 5x^3 - 6x^2 + 24x + 1 &= 0 .\end{aligned}\]Now we can use Vieta: the sum of the roots is $a+b+c=\boxed{\frac65}.$


Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then determine the greater one. Let me think about how to approach this.

First, let me recall that consecutive even integers are numbers like 2 and 4, 6 and 8, etc. They differ by 2. So if I let the first integer be some number, the next one would be that number plus 2.

Let me denote the smaller integer as x. Then the next consecutive even integer would be x + 2. Their product is given as 288. So, the equation would be:

x * (x + 2) = 288

Alright, so expanding that equation:

x² + 2x = 288

To solve this quadratic equation, I need to bring all terms to one side to set the equation equal to zero. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Now, I need to solve this quadratic equation. There are a few methods: factoring, completing the square, or using the quadratic formula. Let me see if this quadratic can be factored.

Looking for two numbers that multiply to -288 and add up to 2. Hmm. Let me think. Since the product is negative, one number has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list the factor pairs of 288:

1 and 288

2 and 144

3 and 96

4 and 72

6 and 48

8 and 36

9 and 32

12 and 24

16 and 18

Now, since we need two numbers that multiply to -288 and add to 2. Let me try some combinations.

Suppose I take 18 and -16. Their product is -288 (since 18 * -16 = -288). Their sum is 18 + (-16) = 2. Yes! That works.

So, the quadratic can be factored as:

(x + 18)(x - 16) = 0

Wait, let me check that. If I expand (x + 18)(x - 16), I get x² -16x + 18x - 288 = x² + 2x - 288. Yes, that's correct.

So, setting each factor equal to zero:

x + 18 = 0 => x = -18

or

x - 16 = 0 => x = 16

But the problem states that the integers are positive. So x = -18 is negative, which we can disregard. Therefore, x = 16 is the smaller integer. Then the next consecutive even integer is 16 + 2 = 18.

Let me check if their product is indeed 288. 16 * 18 = 288. Let me compute that: 16*10=160, 16*8=128, so 160+128=288. Yes, that's correct.

Therefore, the greater of the two integers is 18.

Wait, but just to make sure I didn't make any mistakes in factoring. Let me try solving the quadratic equation using the quadratic formula as well, to cross-verify.

The quadratic equation is x² + 2x - 288 = 0.

Quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a)

Here, a = 1, b = 2, c = -288.

So discriminant D = b² - 4ac = (2)² - 4(1)(-288) = 4 + 1152 = 1156

Square root of 1156 is 34, since 34*34 = 1156.

Therefore, x = [-2 ± 34]/2

So two solutions:

x = (-2 + 34)/2 = 32/2 = 16

x = (-2 - 34)/2 = -36/2 = -18

Same results as before. So x = 16 or x = -18. Since we need positive integers, x = 16. Therefore, the two integers are 16 and 18, so the greater one is 18.

Alternatively, maybe I can think of it another way. Let me try to approximate the square root of 288 to see if that helps.

Since the product of two consecutive even numbers is 288, so if they were consecutive integers (not necessarily even), the numbers would be around sqrt(288). Let me compute sqrt(288). sqrt(289) is 17, so sqrt(288) is just a bit less than 17, say approximately 16.97. So since we are dealing with even integers, the numbers would be around 16 and 18. Which is exactly what we found. So that gives me more confidence.

Alternatively, if I didn't think of factoring, maybe I could have tried plugging in numbers. Let me try that approach as well, just to check.

Let me list even numbers and compute their products with the next even number:

Starting from 2:

2*4=8

4*6=24

6*8=48

8*10=80

10*12=120

12*14=168

14*16=224

16*18=288

Ah, there we go. So 16*18=288. So that's the pair. Therefore, the greater integer is 18. So that's another way to see it.

So all methods lead to the same answer. Therefore, I can be confident that the greater integer is 18.

But just to recap, the key steps were:

1. Let the two consecutive even integers be x and x + 2.

2. Set up the equation x(x + 2) = 288.

3. Expand and rearrange to form a quadratic equation.

4. Solve the quadratic equation by factoring, quadratic formula, or trial and error.

5. Verify the solution by plugging back into the original equation.

All steps confirm that the integers are 16 and 18, so the greater one is 18.

I think that's thorough. I don't see any mistakes in my reasoning. So I think the answer is 18.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we are given that the product of two **consecutive positive even integers** is 288, and we are to find the **greater** of the two integers.

---

### Step 1: Define the Variables

Let the smaller even integer be $ x $. Then the next consecutive even integer is $ x + 2 $.

---

### Step 2: Set Up the Equation

The product of the two integers is:

$$
x(x + 2) = 288
$$

Expanding the left-hand side:

$$
x^2 + 2x = 288
$$

Bring all terms to one side to form a standard quadratic equation:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's try factoring first.

We look for two numbers that multiply to $-288$ and add to $2$. These numbers are $18$ and $-16$:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \quad \text{(discard, since we need positive integers)}
$$
$$
x - 16 = 0 \Rightarrow x = 16
$$

So the two consecutive even integers are:

$$
x = 16 \quad \text{and} \quad x + 2 = 18
$$

---

### Step 4: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms the solution is correct.

---

### Final Answer

The greater of the two integers is:

$$
\boxed{18}
$$

[Assistant]
Explanation:
Okay, so I need to find two consecutive positive even integers whose product is 288, and then determine the greater one. Let me think about how to approach this.

First, let me recall that consecutive even integers are numbers like 2 and 4, 6 and 8, etc. They differ by 2. So if I let the first integer be some number, the next one would be that number plus 2.

Let me denote the smaller integer as x. Then the next consecutive even integer would be x + 2. Their product is given as 288. So, the equation would be:

x * (x + 2) = 288

Alright, so expanding that equation:

x² + 2x = 288

To solve this quadratic equation, I need to bring all terms to one side to set the equation equal to zero. Let me subtract 288 from both sides:

x² + 2x - 288 = 0

Now, I need to solve this quadratic equation. There are a few methods: factoring, completing the square, or using the quadratic formula. Let me see if this quadratic can be factored.

Looking for two numbers that multiply to -288 and add up to 2. Hmm. Let me think. Since the product is negative, one number has to be positive and the other negative. Also, their sum is positive 2. So the positive number must be larger in magnitude than the negative one.

Let me list the factor pairs of 288:

1 and 288

2 and 144

3 and 96

4 and 72

6 and 48

8 and 36

9 and 32

12 and 24

16 and 18

Now, since we need two numbers that multiply to -288 and add to 2. Let me try some combinations.

Suppose I take 18 and -16. Their product is -288 (since 18 * -16 = -288). Their sum is 18 + (-16) = 2. Yes! That works.

So, the quadratic can be factored as:

(x + 18)(x - 16) = 0

Wait, let me check that. If I expand (x + 18)(x - 16), I get x² -16x + 18x - 288 = x² + 2x - 288. Yes, that's correct.

So, setting each factor equal to zero:

x + 18 = 0 => x = -18

or

x - 16 = 0 => x = 16

But the problem states that the integers are positive. So x = -18 is negative, which we can disregard. Therefore, x = 16 is the smaller integer. Then the next consecutive even integer is 16 + 2 = 18.

Let me check if their product is indeed 288. 16 * 18 = 288. Let me compute that: 16*10=160, 16*8=128, so 160+128=288. Yes, that's correct.

Therefore, the greater of the two integers is 18.

Wait, but just to make sure I didn't make any mistakes in factoring. Let me try solving the quadratic equation using the quadratic formula as well, to cross-verify.

The quadratic equation is x² + 2x - 288 = 0.

Quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a)

Here, a = 1, b = 2, c = -288.

So discriminant D = b² - 4ac = (2)² - 4(1)(-288) = 4 + 1152 = 1156

Square root of 1156 is 34, since 34*34 = 1156.

Therefore, x = [-2 ± 34]/2

So two solutions:

x = (-2 + 34)/2 = 32/2 = 16

x = (-2 - 34)/2 = -36/2 = -18

Same results as before. So x = 16 or x = -18. Since we need positive integers, x = 16. Therefore, the two integers are 16 and 18, so the greater one is 18.

Alternatively, maybe I can think of it another way. Let me try to approximate the square root of 288 to see if that helps.

Since the product of two consecutive even numbers is 288, so if they were consecutive integers (not necessarily even), the numbers would be around sqrt(288). Let me compute sqrt(288). sqrt(289) is 17, so sqrt(288) is just a bit less than 17, say approximately 16.97. So since we are dealing with even integers, the numbers would be around 16 and 18. Which is exactly what we found. So that gives me more confidence.

Alternatively, if I didn't think of factoring, maybe I could have tried plugging in numbers. Let me try that approach as well, just to check.

Let me list even numbers and compute their products with the next even number:

Starting from 2:

2*4=8

4*6=24

6*8=48

8*10=80

10*12=120

12*14=168

14*16=224

16*18=288

Ah, there we go. So 16*18=288. So that's the pair. Therefore, the greater integer is 18. So that's another way to see it.

So all methods lead to the same answer. Therefore, I can be confident that the greater integer is 18.

But just to recap, the key steps were:

1. Let the two consecutive even integers be x and x + 2.

2. Set up the equation x(x + 2) = 288.

3. Expand and rearrange to form a quadratic equation.

4. Solve the quadratic equation by factoring, quadratic formula, or trial and error.

5. Verify the solution by plugging back into the original equation.

All steps confirm that the integers are 16 and 18, so the greater one is 18.

I think that's thorough. I don't see any mistakes in my reasoning. So I think the answer is 18.

**Final Answer**
The greater of the two integers is \boxed{18}.

To solve the problem, we are given that the product of two **consecutive positive even integers** is 288, and we are to find the **greater** of the two integers.

---

### Step 1: Define the Variables

Let the smaller even integer be $ x $. Then the next consecutive even integer is $ x + 2 $.

---

### Step 2: Set Up the Equation

The product of the two integers is:

$$
x(x + 2) = 288
$$

Expanding the left-hand side:

$$
x^2 + 2x = 288
$$

Bring all terms to one side to form a standard quadratic equation:

$$
x^2 + 2x - 288 = 0
$$

---

### Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's try factoring first.

We look for two numbers that multiply to $-288$ and add to $2$. These numbers are $18$ and $-16$:

$$
(x + 18)(x - 16) = 0
$$

Setting each factor equal to zero:

$$
x + 18 = 0 \Rightarrow x = -18 \quad \text{(discard, since we need positive integers)}
$$
$$
x - 16 = 0 \Rightarrow x = 16
$$

So the two consecutive even integers are:

$$
x = 16 \quad \text{and} \quad x + 2 = 18
$$

---

### Step 4: Verify the Solution

Check the product:

$$
16 \times 18 = 288
$$

This confirms the solution is correct.

---

### Final Answer

The greater of the two integers is:

$$
\boxed{18}
$$

Solution: First we find the prime factorization of 288 to be $2^5\cdot 3^2$, and we have to split these factors among two consecutive even integers. The 3 must be with at least one 2 for the integer to be even, meaning one of the factors must be a multiple of $6.$ After some playing around, we find that when one factor is 18, that leaves us with $2^4=16$. So, our two integers are 16 and 18, with the greater integer being $\boxed{18}$.


Explanation:
Okay, so I need to evaluate log base 2 of 64. Hmm, let me think. Logarithms can sometimes be tricky, but maybe I can remember what log base 2 means. It's asking the question: 2 raised to what power gives me 64? So, in other words, I need to find an exponent x such that 2^x = 64.

Alright, let me recall the powers of 2. Let me start from the beginning. 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, and then 2^6 should be 64. Wait, is that right? Let me check. 2^5 is 32, so multiplying by 2 again would be 32*2=64. Yes, so 2^6 equals 64. Therefore, if 2^6 = 64, then log base 2 of 64 is 6.

But let me make sure I didn't skip any steps or make a mistake. Maybe I can write out the powers step by step. Let's see:

- 2^1 = 2
- 2^2 = 2*2 = 4
- 2^3 = 4*2 = 8
- 2^4 = 8*2 = 16
- 2^5 = 16*2 = 32
- 2^6 = 32*2 = 64

Yep, that's correct. So, 2 raised to the 6th power is indeed 64. Therefore, the logarithm base 2 of 64 is 6.

Alternatively, maybe I can use the change of base formula to verify. The change of base formula says that log_b(a) = log_c(a) / log_c(b), where c is any positive number. If I choose base 10, for example, then log_2(64) = log10(64)/log10(2). Let me calculate that.

First, log10(64). I know that log10(10) is 1, log10(100) is 2. 64 is between 10 and 100, so log10(64) is between 1 and 2. But I need a more precise value. Wait, maybe I can remember that log10(2) is approximately 0.3010, and log10(64) is log10(2^6) which is 6*log10(2). So that would be 6*0.3010 = 1.806. Then, dividing by log10(2) which is 0.3010, so 1.806 / 0.3010 ≈ 6. So that gives me the same answer, 6.

Alternatively, if I use natural logarithm, ln(64)/ln(2). Again, ln(64) is ln(2^6) = 6*ln(2). Therefore, dividing by ln(2) gives 6. So regardless of the method, I get 6.

Wait, maybe I can think of 64 as a power of 2 in another way? Let me see. 64 is 8 squared, right? 8*8=64. And 8 is 2^3. Therefore, 64 is (2^3)^2 = 2^(3*2) = 2^6. So that's another way to see that 2^6 = 64. Therefore, log base 2 of 64 is 6.

Is there another approach? Maybe using prime factorization? 64 is 2*2*2*2*2*2, which is 2^6. So, yeah, that's the same as before.

Alternatively, if I didn't remember the powers of 2, maybe I can use logarithms properties. For example, log2(64) = log2(2^6) = 6*log2(2) = 6*1 = 6. Oh, that's straightforward. Since 64 is 2^6, then applying the logarithm property that log_b(b^k) = k.

So, all these different methods lead me to the same answer, 6. Therefore, I can be confident that log base 2 of 64 is indeed 6.

Wait, just to make sure I didn't confuse log base 2 with something else. For example, if someone confuses log base 10 or natural log, but the question specifically says log base 2. So, no issues there.

Another check: if I have log2(64) = x, then 2^x = 64. Let me plug in x=6: 2^6=64. Correct. If I plug in x=5, 2^5=32, which is too small. x=7: 2^7=128, which is too big. So, 6 is the correct exponent.

Therefore, after multiple ways of thinking about it and checking, I can confirm that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate the expression $\log_2 64$, we begin by understanding what a logarithm represents. Specifically, $\log_b a = x$ means that $b^x = a$. In this case, we are looking for the exponent $x$ such that:

$$
2^x = 64
$$

---

### Step 1: Express 64 as a power of 2

We know that:

$$
2^1 = 2 \\
2^2 = 4 \\
2^3 = 8 \\
2^4 = 16 \\
2^5 = 32 \\
2^6 = 64
$$

So, $2^6 = 64$. Therefore, the exponent $x$ that satisfies $2^x = 64$ is $x = 6$.

---

### Step 2: Apply the logarithmic identity

Using the logarithmic identity:

$$
\log_b (b^k) = k
$$

We can directly apply this to our case:

$$
\log_2 (2^6) = 6
$$

---

### Step 3: Confirm using change of base formula (optional verification)

As an additional check, we can use the change of base formula:

$$
\log_2 64 = \frac{\log_{10} 64}{\log_{10} 2}
$$

We know that:

$$
64 = 2^6 \Rightarrow \log_{10} 64 = \log_{10} (2^6) = 6 \log_{10} 2
$$

So:

$$
\log_2 64 = \frac{6 \log_{10} 2}{\log_{10} 2} = 6
$$

This confirms our result.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate log base 2 of 64. Hmm, let me think. Logarithms can sometimes be tricky, but maybe I can remember what log base 2 means. It's asking the question: 2 raised to what power gives me 64? So, in other words, I need to find an exponent x such that 2^x = 64.

Alright, let me recall the powers of 2. Let me start from the beginning. 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, and then 2^6 should be 64. Wait, is that right? Let me check. 2^5 is 32, so multiplying by 2 again would be 32*2=64. Yes, so 2^6 equals 64. Therefore, if 2^6 = 64, then log base 2 of 64 is 6.

But let me make sure I didn't skip any steps or make a mistake. Maybe I can write out the powers step by step. Let's see:

- 2^1 = 2
- 2^2 = 2*2 = 4
- 2^3 = 4*2 = 8
- 2^4 = 8*2 = 16
- 2^5 = 16*2 = 32
- 2^6 = 32*2 = 64

Yep, that's correct. So, 2 raised to the 6th power is indeed 64. Therefore, the logarithm base 2 of 64 is 6.

Alternatively, maybe I can use the change of base formula to verify. The change of base formula says that log_b(a) = log_c(a) / log_c(b), where c is any positive number. If I choose base 10, for example, then log_2(64) = log10(64)/log10(2). Let me calculate that.

First, log10(64). I know that log10(10) is 1, log10(100) is 2. 64 is between 10 and 100, so log10(64) is between 1 and 2. But I need a more precise value. Wait, maybe I can remember that log10(2) is approximately 0.3010, and log10(64) is log10(2^6) which is 6*log10(2). So that would be 6*0.3010 = 1.806. Then, dividing by log10(2) which is 0.3010, so 1.806 / 0.3010 ≈ 6. So that gives me the same answer, 6.

Alternatively, if I use natural logarithm, ln(64)/ln(2). Again, ln(64) is ln(2^6) = 6*ln(2). Therefore, dividing by ln(2) gives 6. So regardless of the method, I get 6.

Wait, maybe I can think of 64 as a power of 2 in another way? Let me see. 64 is 8 squared, right? 8*8=64. And 8 is 2^3. Therefore, 64 is (2^3)^2 = 2^(3*2) = 2^6. So that's another way to see that 2^6 = 64. Therefore, log base 2 of 64 is 6.

Is there another approach? Maybe using prime factorization? 64 is 2*2*2*2*2*2, which is 2^6. So, yeah, that's the same as before.

Alternatively, if I didn't remember the powers of 2, maybe I can use logarithms properties. For example, log2(64) = log2(2^6) = 6*log2(2) = 6*1 = 6. Oh, that's straightforward. Since 64 is 2^6, then applying the logarithm property that log_b(b^k) = k.

So, all these different methods lead me to the same answer, 6. Therefore, I can be confident that log base 2 of 64 is indeed 6.

Wait, just to make sure I didn't confuse log base 2 with something else. For example, if someone confuses log base 10 or natural log, but the question specifically says log base 2. So, no issues there.

Another check: if I have log2(64) = x, then 2^x = 64. Let me plug in x=6: 2^6=64. Correct. If I plug in x=5, 2^5=32, which is too small. x=7: 2^7=128, which is too big. So, 6 is the correct exponent.

Therefore, after multiple ways of thinking about it and checking, I can confirm that the answer is 6.

**Final Answer**
The value of $\log_2 64$ is \boxed{6}.

To evaluate the expression $\log_2 64$, we begin by understanding what a logarithm represents. Specifically, $\log_b a = x$ means that $b^x = a$. In this case, we are looking for the exponent $x$ such that:

$$
2^x = 64
$$

---

### Step 1: Express 64 as a power of 2

We know that:

$$
2^1 = 2 \\
2^2 = 4 \\
2^3 = 8 \\
2^4 = 16 \\
2^5 = 32 \\
2^6 = 64
$$

So, $2^6 = 64$. Therefore, the exponent $x$ that satisfies $2^x = 64$ is $x = 6$.

---

### Step 2: Apply the logarithmic identity

Using the logarithmic identity:

$$
\log_b (b^k) = k
$$

We can directly apply this to our case:

$$
\log_2 (2^6) = 6
$$

---

### Step 3: Confirm using change of base formula (optional verification)

As an additional check, we can use the change of base formula:

$$
\log_2 64 = \frac{\log_{10} 64}{\log_{10} 2}
$$

We know that:

$$
64 = 2^6 \Rightarrow \log_{10} 64 = \log_{10} (2^6) = 6 \log_{10} 2
$$

So:

$$
\log_2 64 = \frac{6 \log_{10} 2}{\log_{10} 2} = 6
$$

This confirms our result.

---

### Final Answer

$$
\boxed{6}
$$

Solution: We have $2^6=64$, so $\log_2 64 = \boxed{6}$.


Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, I remember that tan θ is sin θ over cos θ, so maybe I can rewrite the expression in terms of sine and cosine. Let me try that.

So, tan 100° is sin 100° / cos 100°, so the entire expression becomes:

sin 100° / cos 100° + 4 sin 100°

Hmm, maybe I can factor out sin 100°? Let me see:

sin 100° (1 / cos 100° + 4)

But I don't know if that helps. Maybe I need to find a common denominator or combine the terms somehow. Let me try combining them:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Wait, that seems more complicated. Maybe there's another identity I can use here. Let me recall some trigonometric identities.

Alternatively, maybe I can convert degrees to radians? Although I don't know if that helps. Wait, 100 degrees is in the second quadrant, so tan 100° is negative because tangent is negative in the second quadrant. Similarly, sin 100° is positive. So the expression is negative plus positive. Not sure if that helps yet.

Alternatively, maybe I can use some angle addition formulas or something. Let me think. 100° is 60° + 40°, or 90° + 10°, or maybe 180° - 80°. Let me try expressing tan 100° as tan(180° - 80°). Since tan(180° - θ) = -tan θ, so tan 100° = -tan 80°. Similarly, sin 100° = sin(180° - 80°) = sin 80°. So substituting these into the original expression:

-tan 80° + 4 sin 80°

Hmm, that might be helpful. Let me write that down:

- tan 80° + 4 sin 80°

Now, tan 80° is sin 80° / cos 80°, so substituting that:

- (sin 80° / cos 80°) + 4 sin 80°

Again, maybe factor out sin 80°:

sin 80° [ -1 / cos 80° + 4 ]

Which is sin 80° [ 4 - 1 / cos 80° ]

Hmm, not sure if that's helpful. Let me compute the numerical values to see if there's a pattern or if it simplifies to something nice. Maybe I can calculate approximate values?

Let me recall that tan 80° is approximately tan(80°) ≈ 5.671, and sin 80° ≈ 0.9848. Let me compute:

-5.671 + 4 * 0.9848 ≈ -5.671 + 3.939 ≈ -1.732

Wait, -1.732 is approximately -√3, which is about -1.732. So maybe the exact value is -√3? Let me check if that's possible.

So if tan 100° + 4 sin 100° equals -√3? Let me verify with more accurate calculations.

First, let me compute tan 100°. 100 degrees is in the second quadrant. Let me convert 100° to radians to use calculator? Wait, maybe I can use exact values? Hmm, not sure. Alternatively, maybe there's an exact expression.

Alternatively, let me think if there's a way to express this combination as a single trigonometric function. Let me consider writing tan θ + 4 sin θ, where θ is 100°, but maybe there's a way to combine them.

Alternatively, maybe using complex numbers or some other identity? Wait, perhaps using the sine of some angle?

Alternatively, let me think about writing tan θ as sin θ / cos θ, so:

tan θ + 4 sin θ = sin θ / cos θ + 4 sin θ = sin θ (1 / cos θ + 4) = sin θ ( (1 + 4 cos θ) / cos θ )

So that's sin θ (1 + 4 cos θ) / cos θ. Not sure if that helps. Maybe if I can find θ such that 1 + 4 cos θ is related to something?

Alternatively, maybe I can use some substitution. Let me think. Let me consider θ = 100°, so maybe there's a relationship with other angles?

Alternatively, maybe I can use the identity for sin A + sin B or something else? Wait, but here we have tan θ + 4 sin θ. Maybe not directly.

Wait, earlier when I approximated, I got approximately -√3. Let me check with more accurate values.

First, let me compute tan(100°). Let me convert 100° to radians: 100° * π / 180 ≈ 1.74533 radians. The tangent of that is tan(1.74533). Let me recall that tan(π/2) is undefined, and π/2 is about 1.5708, so 1.74533 is π/2 + 0.1745 radians, which is about 10 degrees. So tan(π/2 + α) = -cot α. So tan(100°) = tan(90° + 10°) = -cot 10°. Cot 10° is 1/tan 10° ≈ 1 / 0.1763 ≈ 5.671, so tan 100° ≈ -5.671. Then sin 100° is sin(90° + 10°) = cos 10° ≈ 0.9848. Therefore, 4 sin 100° ≈ 4 * 0.9848 ≈ 3.939. Then tan 100° + 4 sin 100° ≈ -5.671 + 3.939 ≈ -1.732, which is approximately -√3 ≈ -1.732. So that suggests that the exact value is -√3. Therefore, maybe the answer is -√3. But how to show that?

So, if I can show that tan 100° + 4 sin 100° = -√3, then that would be the answer. Let me try to prove that.

Let me write tan 100° + 4 sin 100° = -√3. Let me use the earlier substitution where tan 100° = -tan 80°, and sin 100° = sin 80°, so:

- tan 80° + 4 sin 80° = -√3

Let me compute tan 80° and sin 80°. Let me recall that tan 80° = sin 80° / cos 80°, so:

- (sin 80° / cos 80°) + 4 sin 80° = sin 80° [ -1 / cos 80° + 4 ] = sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ]

So, if this equals -√3, then:

sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ] = -√3

Let me compute the numerator inside the brackets: -1 + 4 cos 80°. Let me compute cos 80°. Cos 80° is approximately 0.1736. So 4 * 0.1736 ≈ 0.6944. Then -1 + 0.6944 ≈ -0.3056. Then sin 80° ≈ 0.9848. So multiplying them: 0.9848 * (-0.3056) / 0.1736 ≈ ?

Wait, let me do that step by step. The entire expression is:

sin 80° * (-1 + 4 cos 80°) / cos 80° ≈ 0.9848 * (-0.3056) / 0.1736 ≈ 0.9848 * (-1.76) ≈ -1.732, which is -√3. So that seems to check out numerically. Therefore, the exact value is -√3. But how to show that algebraically?

Let me try to find exact expressions. Let me consider that 80° is related to 30°, 60°, etc. Maybe using some identities. Let me think. Let me recall that cos 80° can be written in terms of sin 10°, since cos 80° = sin 10°, and sin 80° = cos 10°. Maybe that helps?

Alternatively, let me consider using the sine of some angle. Let me think. Suppose I want to show that:

sin 80° [ -1 + 4 cos 80° ] / cos 80° = -√3

Let me rearrange:

sin 80° [ -1 + 4 cos 80° ] = -√3 cos 80°

Let me compute the left-hand side:

- sin 80° + 4 sin 80° cos 80°

Hmm, 4 sin 80° cos 80° can be written as 2 * 2 sin 80° cos 80° = 2 sin 160°, since sin 2θ = 2 sin θ cos θ. Therefore:

Left-hand side becomes - sin 80° + 2 sin 160°

But sin 160° is sin(180° - 20°) = sin 20°, so:

Left-hand side is - sin 80° + 2 sin 20°

So we have:

- sin 80° + 2 sin 20° = -√3 cos 80° ?

Let me check if this equality holds. Let me compute both sides numerically.

Left-hand side: - sin 80° + 2 sin 20° ≈ -0.9848 + 2 * 0.3420 ≈ -0.9848 + 0.6840 ≈ -0.3008

Right-hand side: -√3 cos 80° ≈ -1.732 * 0.1736 ≈ -0.3008

Yes! They are equal. Therefore, the equality holds. Therefore, the original expression simplifies to -√3. Therefore, tan 100° + 4 sin 100° = -√3.

But let me verify this step-by-step algebraically. Let me write:

Starting from:

- sin 80° + 2 sin 20° = -√3 cos 80° ?

Let me rearrange:

- sin 80° + 2 sin 20° + √3 cos 80° = 0 ?

Is this identity true? Let me try to verify it.

Alternatively, maybe express sin 80° and sin 20° in terms of other angles. Let me recall that 80° = 60° + 20°, so maybe use sine addition formula.

Let me compute sin 80° = sin(60° + 20°) = sin 60° cos 20° + cos 60° sin 20° = (√3/2) cos 20° + (1/2) sin 20°

Therefore, substituting into the left-hand side:

- [ (√3/2) cos 20° + (1/2) sin 20° ] + 2 sin 20° = - (√3/2) cos 20° - (1/2) sin 20° + 2 sin 20° = - (√3/2) cos 20° + (3/2) sin 20°

So we have:

- (√3/2) cos 20° + (3/2) sin 20° = ?

Let me factor out 1/2:

(1/2)[ -√3 cos 20° + 3 sin 20° ]

Now, let me see if this equals -√3 cos 80°. Let me compute the right-hand side:

-√3 cos 80° = -√3 sin 10°, since cos 80° = sin 10°. But not sure if that helps. Alternatively, let me compute the expression:

(1/2)[ -√3 cos 20° + 3 sin 20° ]

Let me think if this can be expressed as a single sine or cosine function. Let me recall that expressions of the form A sin θ + B cos θ can be written as C sin(θ + φ) or something like that. Let me try that.

Let me write -√3 cos 20° + 3 sin 20° as R sin(20° + φ) or R cos(20° + φ). Let me try to write it as R sin(20° + φ). Let me recall that:

R sin(θ + φ) = R sin θ cos φ + R cos θ sin φ

Comparing with -√3 cos 20° + 3 sin 20°, which is 3 sin 20° - √3 cos 20°. So:

R sin θ cos φ + R cos θ sin φ = 3 sin 20° - √3 cos 20°

So, if we set θ = 20°, then:

R cos φ = 3

R sin φ = -√3

Therefore, dividing the two equations:

tan φ = (-√3)/3 = -1/√3 => φ = -30°, since tan(-30°) = -1/√3.

Then R^2 = (R cos φ)^2 + (R sin φ)^2 = 3^2 + (-√3)^2 = 9 + 3 = 12 => R = 2√3.

Therefore, the expression becomes:

2√3 sin(20° - 30°) = 2√3 sin(-10°) = -2√3 sin 10°

Therefore, going back:

(1/2)[ -√3 cos 20° + 3 sin 20° ] = (1/2)( -2√3 sin 10° ) = -√3 sin 10°

But earlier, we had that the left-hand side equals -√3 cos 80°, and since cos 80° = sin 10°, this is indeed equal. Therefore, we have:

Left-hand side equals -√3 sin 10° = -√3 cos 80°, which matches the right-hand side. Therefore, the equality holds. Therefore, our previous steps are correct, and the original expression simplifies to -√3.

Therefore, tan 100° + 4 sin 100° = -√3.

Let me recap the steps to make sure I didn't skip anything:

1. Recognize that tan 100° = -tan 80° and sin 100° = sin 80°, so the expression becomes -tan 80° + 4 sin 80°.

2. Express tan 80° as sin 80° / cos 80°, leading to -sin 80° / cos 80° + 4 sin 80°.

3. Factor out sin 80°, resulting in sin 80° [ -1 / cos 80° + 4 ].

4. Convert the expression inside the brackets to combine terms, leading to sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ].

5. Multiply out to get -sin 80° + 4 sin 80° cos 80°, then use double-angle identity to write 4 sin 80° cos 80° as 2 sin 160°, which is 2 sin 20°.

6. Substitute sin 160° = sin 20°, leading to -sin 80° + 2 sin 20°.

7. Use angle addition formula to express sin 80° as sin(60° + 20°), expand, and simplify the expression to -√3 sin 10°, which is equal to -√3 cos 80°, confirming the equality.

Therefore, all steps check out, and the final result is indeed -√3.

Alternatively, another approach could be using complex numbers or other identities, but this method seems solid. Let me just check once more with exact values.

Alternatively, suppose I consider the original expression tan 100° + 4 sin 100°, and I want to relate it to known angles. Since 100° is 60° + 40°, but not sure. Alternatively, maybe using the sine of 100° and tangent of 100° in terms of known triangles? Not sure. But since we already have a numerical confirmation and an algebraic proof, I think we can be confident.

Therefore, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by recognizing that $100^\circ$ lies in the second quadrant, where:

- $\tan \theta < 0$
- $\sin \theta > 0$

This gives us a hint that the expression will result in a negative value. Let's proceed with a trigonometric identity to simplify the expression.

---

### Step 1: Use Trigonometric Identities

We use the identities:

$$
\tan(180^\circ - \theta) = -\tan \theta \quad \text{and} \quad \sin(180^\circ - \theta) = \sin \theta
$$

So:

$$
\tan 100^\circ = \tan(180^\circ - 80^\circ) = -\tan 80^\circ
$$
$$
\sin 100^\circ = \sin(180^\circ - 80^\circ) = \sin 80^\circ
$$

Substituting into the original expression:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

---

### Step 2: Express in Terms of Sine and Cosine

Recall that:

$$
\tan 80^\circ = \frac{\sin 80^\circ}{\cos 80^\circ}
$$

So the expression becomes:

$$
-\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

Factor out $\sin 80^\circ$:

$$
\sin 80^\circ \left( -\frac{1}{\cos 80^\circ} + 4 \right)
$$

Now combine the terms in the parentheses:

$$
\sin 80^\circ \left( \frac{-1 + 4 \cos 80^\circ}{\cos 80^\circ} \right)
$$

---

### Step 3: Use Trigonometric Identities to Simplify

We now simplify the numerator:

$$
-1 + 4 \cos 80^\circ
$$

Let’s use the identity:

$$
2 \sin A \cos A = \sin 2A
$$

So:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \cdot 2 \sin 80^\circ \cos 80^\circ = 2 \sin 160^\circ
$$

But $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$, so:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \sin 20^\circ
$$

Now, the expression becomes:

$$
- \sin 80^\circ + 2 \sin 20^\circ
$$

---

### Step 4: Use Angle Addition Formula

We use the identity:

$$
\sin(60^\circ + 20^\circ) = \sin 80^\circ = \sin 60^\circ \cos 20^\circ + \cos 60^\circ \sin 20^\circ
$$

$$
= \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ
$$

Substitute into the expression:

$$
- \left( \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ \right) + 2 \sin 20^\circ
$$

$$
= -\frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ + 2 \sin 20^\circ
$$

$$
= -\frac{\sqrt{3}}{2} \cos 20^\circ + \frac{3}{2} \sin 20^\circ
$$

Factor out $\frac{1}{2}$:

$$
\frac{1}{2} \left( -\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ \right)
$$

Now, we recognize this as a form of a sine function:

$$
A \sin \theta + B \cos \theta = R \sin(\theta + \phi)
$$

Let’s write:

$$
-\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ = R \sin(20^\circ + \phi)
$$

We find:

- $R = \sqrt{(-\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$
- $\tan \phi = \frac{-\sqrt{3}}{3} = -\frac{1}{\sqrt{3}} \Rightarrow \phi = -30^\circ$

So:

$$
-\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ = 2\sqrt{3} \sin(20^\circ - 30^\circ) = 2\sqrt{3} \sin(-10^\circ) = -2\sqrt{3} \sin 10^\circ
$$

Therefore, the entire expression becomes:

$$
\frac{1}{2} \cdot (-2\sqrt{3} \sin 10^\circ) = -\sqrt{3} \sin 10^\circ
$$

But $\sin 10^\circ = \cos 80^\circ$, so:

$$
-\sqrt{3} \sin 10^\circ = -\sqrt{3} \cos 80^\circ
$$

---

### Final Step: Conclusion

We have shown that:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer

$$
\boxed{-\sqrt{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the expression tan 100° + 4 sin 100°. Hmm, let me think about how to approach this. First, I remember that tan θ is sin θ over cos θ, so maybe I can rewrite the expression in terms of sine and cosine. Let me try that.

So, tan 100° is sin 100° / cos 100°, so the entire expression becomes:

sin 100° / cos 100° + 4 sin 100°

Hmm, maybe I can factor out sin 100°? Let me see:

sin 100° (1 / cos 100° + 4)

But I don't know if that helps. Maybe I need to find a common denominator or combine the terms somehow. Let me try combining them:

[sin 100° + 4 sin 100° cos 100°] / cos 100°

Wait, that seems more complicated. Maybe there's another identity I can use here. Let me recall some trigonometric identities.

Alternatively, maybe I can convert degrees to radians? Although I don't know if that helps. Wait, 100 degrees is in the second quadrant, so tan 100° is negative because tangent is negative in the second quadrant. Similarly, sin 100° is positive. So the expression is negative plus positive. Not sure if that helps yet.

Alternatively, maybe I can use some angle addition formulas or something. Let me think. 100° is 60° + 40°, or 90° + 10°, or maybe 180° - 80°. Let me try expressing tan 100° as tan(180° - 80°). Since tan(180° - θ) = -tan θ, so tan 100° = -tan 80°. Similarly, sin 100° = sin(180° - 80°) = sin 80°. So substituting these into the original expression:

-tan 80° + 4 sin 80°

Hmm, that might be helpful. Let me write that down:

- tan 80° + 4 sin 80°

Now, tan 80° is sin 80° / cos 80°, so substituting that:

- (sin 80° / cos 80°) + 4 sin 80°

Again, maybe factor out sin 80°:

sin 80° [ -1 / cos 80° + 4 ]

Which is sin 80° [ 4 - 1 / cos 80° ]

Hmm, not sure if that's helpful. Let me compute the numerical values to see if there's a pattern or if it simplifies to something nice. Maybe I can calculate approximate values?

Let me recall that tan 80° is approximately tan(80°) ≈ 5.671, and sin 80° ≈ 0.9848. Let me compute:

-5.671 + 4 * 0.9848 ≈ -5.671 + 3.939 ≈ -1.732

Wait, -1.732 is approximately -√3, which is about -1.732. So maybe the exact value is -√3? Let me check if that's possible.

So if tan 100° + 4 sin 100° equals -√3? Let me verify with more accurate calculations.

First, let me compute tan 100°. 100 degrees is in the second quadrant. Let me convert 100° to radians to use calculator? Wait, maybe I can use exact values? Hmm, not sure. Alternatively, maybe there's an exact expression.

Alternatively, let me think if there's a way to express this combination as a single trigonometric function. Let me consider writing tan θ + 4 sin θ, where θ is 100°, but maybe there's a way to combine them.

Alternatively, maybe using complex numbers or some other identity? Wait, perhaps using the sine of some angle?

Alternatively, let me think about writing tan θ as sin θ / cos θ, so:

tan θ + 4 sin θ = sin θ / cos θ + 4 sin θ = sin θ (1 / cos θ + 4) = sin θ ( (1 + 4 cos θ) / cos θ )

So that's sin θ (1 + 4 cos θ) / cos θ. Not sure if that helps. Maybe if I can find θ such that 1 + 4 cos θ is related to something?

Alternatively, maybe I can use some substitution. Let me think. Let me consider θ = 100°, so maybe there's a relationship with other angles?

Alternatively, maybe I can use the identity for sin A + sin B or something else? Wait, but here we have tan θ + 4 sin θ. Maybe not directly.

Wait, earlier when I approximated, I got approximately -√3. Let me check with more accurate values.

First, let me compute tan(100°). Let me convert 100° to radians: 100° * π / 180 ≈ 1.74533 radians. The tangent of that is tan(1.74533). Let me recall that tan(π/2) is undefined, and π/2 is about 1.5708, so 1.74533 is π/2 + 0.1745 radians, which is about 10 degrees. So tan(π/2 + α) = -cot α. So tan(100°) = tan(90° + 10°) = -cot 10°. Cot 10° is 1/tan 10° ≈ 1 / 0.1763 ≈ 5.671, so tan 100° ≈ -5.671. Then sin 100° is sin(90° + 10°) = cos 10° ≈ 0.9848. Therefore, 4 sin 100° ≈ 4 * 0.9848 ≈ 3.939. Then tan 100° + 4 sin 100° ≈ -5.671 + 3.939 ≈ -1.732, which is approximately -√3 ≈ -1.732. So that suggests that the exact value is -√3. Therefore, maybe the answer is -√3. But how to show that?

So, if I can show that tan 100° + 4 sin 100° = -√3, then that would be the answer. Let me try to prove that.

Let me write tan 100° + 4 sin 100° = -√3. Let me use the earlier substitution where tan 100° = -tan 80°, and sin 100° = sin 80°, so:

- tan 80° + 4 sin 80° = -√3

Let me compute tan 80° and sin 80°. Let me recall that tan 80° = sin 80° / cos 80°, so:

- (sin 80° / cos 80°) + 4 sin 80° = sin 80° [ -1 / cos 80° + 4 ] = sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ]

So, if this equals -√3, then:

sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ] = -√3

Let me compute the numerator inside the brackets: -1 + 4 cos 80°. Let me compute cos 80°. Cos 80° is approximately 0.1736. So 4 * 0.1736 ≈ 0.6944. Then -1 + 0.6944 ≈ -0.3056. Then sin 80° ≈ 0.9848. So multiplying them: 0.9848 * (-0.3056) / 0.1736 ≈ ?

Wait, let me do that step by step. The entire expression is:

sin 80° * (-1 + 4 cos 80°) / cos 80° ≈ 0.9848 * (-0.3056) / 0.1736 ≈ 0.9848 * (-1.76) ≈ -1.732, which is -√3. So that seems to check out numerically. Therefore, the exact value is -√3. But how to show that algebraically?

Let me try to find exact expressions. Let me consider that 80° is related to 30°, 60°, etc. Maybe using some identities. Let me think. Let me recall that cos 80° can be written in terms of sin 10°, since cos 80° = sin 10°, and sin 80° = cos 10°. Maybe that helps?

Alternatively, let me consider using the sine of some angle. Let me think. Suppose I want to show that:

sin 80° [ -1 + 4 cos 80° ] / cos 80° = -√3

Let me rearrange:

sin 80° [ -1 + 4 cos 80° ] = -√3 cos 80°

Let me compute the left-hand side:

- sin 80° + 4 sin 80° cos 80°

Hmm, 4 sin 80° cos 80° can be written as 2 * 2 sin 80° cos 80° = 2 sin 160°, since sin 2θ = 2 sin θ cos θ. Therefore:

Left-hand side becomes - sin 80° + 2 sin 160°

But sin 160° is sin(180° - 20°) = sin 20°, so:

Left-hand side is - sin 80° + 2 sin 20°

So we have:

- sin 80° + 2 sin 20° = -√3 cos 80° ?

Let me check if this equality holds. Let me compute both sides numerically.

Left-hand side: - sin 80° + 2 sin 20° ≈ -0.9848 + 2 * 0.3420 ≈ -0.9848 + 0.6840 ≈ -0.3008

Right-hand side: -√3 cos 80° ≈ -1.732 * 0.1736 ≈ -0.3008

Yes! They are equal. Therefore, the equality holds. Therefore, the original expression simplifies to -√3. Therefore, tan 100° + 4 sin 100° = -√3.

But let me verify this step-by-step algebraically. Let me write:

Starting from:

- sin 80° + 2 sin 20° = -√3 cos 80° ?

Let me rearrange:

- sin 80° + 2 sin 20° + √3 cos 80° = 0 ?

Is this identity true? Let me try to verify it.

Alternatively, maybe express sin 80° and sin 20° in terms of other angles. Let me recall that 80° = 60° + 20°, so maybe use sine addition formula.

Let me compute sin 80° = sin(60° + 20°) = sin 60° cos 20° + cos 60° sin 20° = (√3/2) cos 20° + (1/2) sin 20°

Therefore, substituting into the left-hand side:

- [ (√3/2) cos 20° + (1/2) sin 20° ] + 2 sin 20° = - (√3/2) cos 20° - (1/2) sin 20° + 2 sin 20° = - (√3/2) cos 20° + (3/2) sin 20°

So we have:

- (√3/2) cos 20° + (3/2) sin 20° = ?

Let me factor out 1/2:

(1/2)[ -√3 cos 20° + 3 sin 20° ]

Now, let me see if this equals -√3 cos 80°. Let me compute the right-hand side:

-√3 cos 80° = -√3 sin 10°, since cos 80° = sin 10°. But not sure if that helps. Alternatively, let me compute the expression:

(1/2)[ -√3 cos 20° + 3 sin 20° ]

Let me think if this can be expressed as a single sine or cosine function. Let me recall that expressions of the form A sin θ + B cos θ can be written as C sin(θ + φ) or something like that. Let me try that.

Let me write -√3 cos 20° + 3 sin 20° as R sin(20° + φ) or R cos(20° + φ). Let me try to write it as R sin(20° + φ). Let me recall that:

R sin(θ + φ) = R sin θ cos φ + R cos θ sin φ

Comparing with -√3 cos 20° + 3 sin 20°, which is 3 sin 20° - √3 cos 20°. So:

R sin θ cos φ + R cos θ sin φ = 3 sin 20° - √3 cos 20°

So, if we set θ = 20°, then:

R cos φ = 3

R sin φ = -√3

Therefore, dividing the two equations:

tan φ = (-√3)/3 = -1/√3 => φ = -30°, since tan(-30°) = -1/√3.

Then R^2 = (R cos φ)^2 + (R sin φ)^2 = 3^2 + (-√3)^2 = 9 + 3 = 12 => R = 2√3.

Therefore, the expression becomes:

2√3 sin(20° - 30°) = 2√3 sin(-10°) = -2√3 sin 10°

Therefore, going back:

(1/2)[ -√3 cos 20° + 3 sin 20° ] = (1/2)( -2√3 sin 10° ) = -√3 sin 10°

But earlier, we had that the left-hand side equals -√3 cos 80°, and since cos 80° = sin 10°, this is indeed equal. Therefore, we have:

Left-hand side equals -√3 sin 10° = -√3 cos 80°, which matches the right-hand side. Therefore, the equality holds. Therefore, our previous steps are correct, and the original expression simplifies to -√3.

Therefore, tan 100° + 4 sin 100° = -√3.

Let me recap the steps to make sure I didn't skip anything:

1. Recognize that tan 100° = -tan 80° and sin 100° = sin 80°, so the expression becomes -tan 80° + 4 sin 80°.

2. Express tan 80° as sin 80° / cos 80°, leading to -sin 80° / cos 80° + 4 sin 80°.

3. Factor out sin 80°, resulting in sin 80° [ -1 / cos 80° + 4 ].

4. Convert the expression inside the brackets to combine terms, leading to sin 80° [ ( -1 + 4 cos 80° ) / cos 80° ].

5. Multiply out to get -sin 80° + 4 sin 80° cos 80°, then use double-angle identity to write 4 sin 80° cos 80° as 2 sin 160°, which is 2 sin 20°.

6. Substitute sin 160° = sin 20°, leading to -sin 80° + 2 sin 20°.

7. Use angle addition formula to express sin 80° as sin(60° + 20°), expand, and simplify the expression to -√3 sin 10°, which is equal to -√3 cos 80°, confirming the equality.

Therefore, all steps check out, and the final result is indeed -√3.

Alternatively, another approach could be using complex numbers or other identities, but this method seems solid. Let me just check once more with exact values.

Alternatively, suppose I consider the original expression tan 100° + 4 sin 100°, and I want to relate it to known angles. Since 100° is 60° + 40°, but not sure. Alternatively, maybe using the sine of 100° and tangent of 100° in terms of known triangles? Not sure. But since we already have a numerical confirmation and an algebraic proof, I think we can be confident.

Therefore, the simplified form of tan 100° + 4 sin 100° is -√3.

**Final Answer**
\boxed{-\sqrt{3}}

To simplify the expression:

$$
\tan 100^\circ + 4 \sin 100^\circ
$$

we begin by recognizing that $100^\circ$ lies in the second quadrant, where:

- $\tan \theta < 0$
- $\sin \theta > 0$

This gives us a hint that the expression will result in a negative value. Let's proceed with a trigonometric identity to simplify the expression.

---

### Step 1: Use Trigonometric Identities

We use the identities:

$$
\tan(180^\circ - \theta) = -\tan \theta \quad \text{and} \quad \sin(180^\circ - \theta) = \sin \theta
$$

So:

$$
\tan 100^\circ = \tan(180^\circ - 80^\circ) = -\tan 80^\circ
$$
$$
\sin 100^\circ = \sin(180^\circ - 80^\circ) = \sin 80^\circ
$$

Substituting into the original expression:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\tan 80^\circ + 4 \sin 80^\circ
$$

---

### Step 2: Express in Terms of Sine and Cosine

Recall that:

$$
\tan 80^\circ = \frac{\sin 80^\circ}{\cos 80^\circ}
$$

So the expression becomes:

$$
-\frac{\sin 80^\circ}{\cos 80^\circ} + 4 \sin 80^\circ
$$

Factor out $\sin 80^\circ$:

$$
\sin 80^\circ \left( -\frac{1}{\cos 80^\circ} + 4 \right)
$$

Now combine the terms in the parentheses:

$$
\sin 80^\circ \left( \frac{-1 + 4 \cos 80^\circ}{\cos 80^\circ} \right)
$$

---

### Step 3: Use Trigonometric Identities to Simplify

We now simplify the numerator:

$$
-1 + 4 \cos 80^\circ
$$

Let’s use the identity:

$$
2 \sin A \cos A = \sin 2A
$$

So:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \cdot 2 \sin 80^\circ \cos 80^\circ = 2 \sin 160^\circ
$$

But $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$, so:

$$
4 \sin 80^\circ \cos 80^\circ = 2 \sin 20^\circ
$$

Now, the expression becomes:

$$
- \sin 80^\circ + 2 \sin 20^\circ
$$

---

### Step 4: Use Angle Addition Formula

We use the identity:

$$
\sin(60^\circ + 20^\circ) = \sin 80^\circ = \sin 60^\circ \cos 20^\circ + \cos 60^\circ \sin 20^\circ
$$

$$
= \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ
$$

Substitute into the expression:

$$
- \left( \frac{\sqrt{3}}{2} \cos 20^\circ + \frac{1}{2} \sin 20^\circ \right) + 2 \sin 20^\circ
$$

$$
= -\frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ + 2 \sin 20^\circ
$$

$$
= -\frac{\sqrt{3}}{2} \cos 20^\circ + \frac{3}{2} \sin 20^\circ
$$

Factor out $\frac{1}{2}$:

$$
\frac{1}{2} \left( -\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ \right)
$$

Now, we recognize this as a form of a sine function:

$$
A \sin \theta + B \cos \theta = R \sin(\theta + \phi)
$$

Let’s write:

$$
-\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ = R \sin(20^\circ + \phi)
$$

We find:

- $R = \sqrt{(-\sqrt{3})^2 + 3^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$
- $\tan \phi = \frac{-\sqrt{3}}{3} = -\frac{1}{\sqrt{3}} \Rightarrow \phi = -30^\circ$

So:

$$
-\sqrt{3} \cos 20^\circ + 3 \sin 20^\circ = 2\sqrt{3} \sin(20^\circ - 30^\circ) = 2\sqrt{3} \sin(-10^\circ) = -2\sqrt{3} \sin 10^\circ
$$

Therefore, the entire expression becomes:

$$
\frac{1}{2} \cdot (-2\sqrt{3} \sin 10^\circ) = -\sqrt{3} \sin 10^\circ
$$

But $\sin 10^\circ = \cos 80^\circ$, so:

$$
-\sqrt{3} \sin 10^\circ = -\sqrt{3} \cos 80^\circ
$$

---

### Final Step: Conclusion

We have shown that:

$$
\tan 100^\circ + 4 \sin 100^\circ = -\sqrt{3}
$$

---

### Final Answer

$$
\boxed{-\sqrt{3}}
$$

Solution:
We have that
\begin{align*}
\tan 100^\circ + 4 \sin 100^\circ &= \frac{\sin 100^\circ}{\cos 100^\circ} + 4 \sin 100^\circ \\
&= \frac{\sin 80^\circ}{-\cos 80^\circ} + 4 \sin 80^\circ \\
&= -\frac{\cos 10^\circ}{\sin 10^\circ} + 4 \cos 10^\circ \\
&= \frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ}.
\end{align*}By double angle formula,
\begin{align*}
\frac{4 \cos 10^\circ \sin 10^\circ - \cos 10^\circ}{\sin 10^\circ} &= \frac{2 \sin 20^\circ - \cos 10^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 80^\circ = 2 \cos 50^\circ \sin (-30^\circ) = -\cos 50^\circ,\]so
\begin{align*}
\frac{\sin 20^\circ + \sin 20^\circ - \sin 80^\circ}{\sin 10^\circ} &= \frac{\sin 20^\circ - \cos 50^\circ}{\sin 10^\circ} \\
&= \frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ}.
\end{align*}By sum-to-product,
\[\sin 20^\circ - \sin 40^\circ = 2 \cos 30^\circ \sin (-10^\circ) = -\sqrt{3} \sin 10^\circ,\]so $\frac{\sin 20^\circ - \sin 40^\circ}{\sin 10^\circ} = \boxed{-\sqrt{3}}.$


Explanation:
Okay, so I need to find the area of a triangle with sides 5, 5, and 6 units. Hmm, let me think. First, since two sides are equal (both 5 units), this must be an isosceles triangle. I remember there are different ways to calculate the area of a triangle. The most common one is the base times height divided by two, right? But I don't know the height here. Maybe I can figure it out?

Alternatively, there's Heron's formula, which uses the lengths of all three sides. Let me recall how that works. Heron's formula says that if a triangle has sides of length a, b, and c, then the area is the square root of [s(s - a)(s - b)(s - c)], where s is the semi-perimeter. Let me try that.

First, let me calculate the semi-perimeter. The perimeter is 5 + 5 + 6, which is 16. So the semi-perimeter s would be 16 divided by 2, which is 8. Okay, so s = 8.

Now plugging into Heron's formula: Area = sqrt[s(s - a)(s - b)(s - c)]. Let me substitute the values. So that would be sqrt[8*(8 - 5)*(8 - 5)*(8 - 6)]. Let me compute each term:

s - a = 8 - 5 = 3

s - b = 8 - 5 = 3 (since b is also 5)

s - c = 8 - 6 = 2

So multiplying all together: 8 * 3 * 3 * 2. Let me compute that step by step. 8*3 is 24, then 24*3 is 72, then 72*2 is 144. So the area is sqrt(144). The square root of 144 is 12. So the area is 12 square units? Wait, that seems straightforward. Let me check if I did that right.

Alternatively, maybe I can use the base and height method. Let me try that to verify. Let's take the base as 6 units. Then, since it's an isosceles triangle, the height can be found by splitting the base into two equal parts. So the base is 6, so each half is 3 units. Then, the height forms a right triangle with the side of 5 units and the base of 3 units. So using Pythagoras: height squared + 3 squared = 5 squared. So height squared = 25 - 9 = 16. Therefore, height is 4. Then area is (base * height)/2 = (6 * 4)/2 = 24/2 = 12. Okay, same answer. So that confirms it. So the area is 12 square units.

Wait, so both methods give me 12. That seems solid. Let me just think if there's another way to check. Maybe using coordinates? Let me try that.

Let me place the triangle on a coordinate system. Let me put the base of 6 units along the x-axis, from (0, 0) to (6, 0). The third vertex is somewhere above the base. Since it's an isosceles triangle with sides 5, 5, 6, the third vertex should be centered above the base. Wait, the base is 6 units, so the midpoint is at (3, 0). The height we found earlier is 4, so the third vertex is at (3, 4). Then, the three vertices are (0, 0), (6, 0), and (3, 4). Let me compute the area using coordinates. There's the shoelace formula, which is another way.

Shoelace formula: For coordinates (x1, y1), (x2, y2), (x3, y3), the area is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|.

Plugging in the points:

x1 = 0, y1 = 0

x2 = 6, y2 = 0

x3 = 3, y3 = 4

So compute:

0*(0 - 4) + 6*(4 - 0) + 3*(0 - 0) all over 2, absolute value.

Calculates to:

0 + 6*4 + 3*0 = 24. Then divide by 2: 24/2 = 12. Absolute value is 12. So again, same answer. So that's three different methods giving me 12. That seems pretty confident.

Alternatively, maybe I can use vectors or trigonometry? Let me try trigonometry. If I can find an angle, then use the formula (1/2)*ab*sin(theta). Let me see. Let me consider the two sides of 5 units and the included angle. Wait, if I can find the angle between the two equal sides, then I can compute the area.

But how do I find that angle? Well, using the Law of Cosines. For a triangle with sides a, b, c, c² = a² + b² - 2ab*cos(theta), where theta is the angle between sides a and b.

In this case, let me take the two sides of 5 units and the included angle. The third side is 6 units. So:

6² = 5² + 5² - 2*5*5*cos(theta)

So 36 = 25 + 25 - 50*cos(theta)

Simplify: 36 = 50 - 50*cos(theta)

Subtract 50: 36 - 50 = -50*cos(theta)

-14 = -50*cos(theta)

Divide both sides by -50: cos(theta) = 14/50 = 7/25

So theta = arccos(7/25). Then, the area is (1/2)*ab*sin(theta) = (1/2)*5*5*sin(theta) = (25/2)*sin(theta)

Now, I need to find sin(theta). Since cos(theta) = 7/25, then sin(theta) can be found using sin²(theta) + cos²(theta) = 1. Therefore, sin(theta) = sqrt(1 - (7/25)^2) = sqrt(1 - 49/625) = sqrt(576/625) = 24/25. Since theta is an angle in a triangle, it's between 0 and 180 degrees, so sine is positive.

Therefore, area is (25/2)*(24/25) = (24/2) = 12. Again, same answer. Wow, so all methods lead to 12. That's reassuring.

So, I think the area is definitely 12 square units. I don't see any mistakes in my calculations. All different approaches confirm the same result. So I can be confident that the answer is 12.

**Final Answer**
The area of the triangle is \boxed{12} square units.

To find the area of a triangle with sides measuring 5, 5, and 6 units, we can use multiple approaches. Here's a step-by-step breakdown of the most effective method, followed by verification using alternative techniques.

---

### **Step 1: Use Heron's Formula**

Heron's formula is a reliable method for calculating the area of a triangle when all three side lengths are known. The formula is:

$$
\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}
$$

Where:
- $ a = 5 $, $ b = 5 $, $ c = 6 $
- $ s $ is the semi-perimeter:
  $$
  s = \frac{a + b + c}{2} = \frac{5 + 5 + 6}{2} = 8
  $$

Now plug into the formula:

$$
\text{Area} = \sqrt{8(8 - 5)(8 - 5)(8 - 6)} = \sqrt{8 \cdot 3 \cdot 3 \cdot 2}
$$

$$
= \sqrt{144} = 12
$$

---

### **Step 2: Verify Using Base and Height**

Since the triangle is isosceles (two sides of 5 units), we can drop a perpendicular from the vertex opposite the base (6 units) to the base, splitting it into two equal segments of 3 units each. This forms two right triangles with:

- Hypotenuse = 5 units
- One leg = 3 units
- The other leg = height $ h $

Using the Pythagorean Theorem:

$$
h^2 + 3^2 = 5^2 \Rightarrow h^2 = 25 - 9 = 16 \Rightarrow h = 4
$$

Now compute the area:

$$
\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 4 = 12
$$

---

### **Step 3: Confirm with Coordinate Geometry (Shoelace Formula)**

Place the triangle on the coordinate plane with vertices at:
- $ A = (0, 0) $
- $ B = (6, 0) $
- $ C = (3, 4) $

Using the shoelace formula:

$$
\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|
$$

$$
= \frac{1}{2} |0(0 - 4) + 6(4 - 0) + 3(0 - 0)| = \frac{1}{2} |0 + 24 + 0| = \frac{1}{2} \cdot 24 = 12
$$

---

### **Step 4: Use Trigonometry (Law of Cosines and Sine)**

Using the Law of Cosines to find the angle between the two equal sides:

$$
6^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \cos(\theta)
\Rightarrow 36 = 25 + 25 - 50\cos(\theta)
\Rightarrow \cos(\theta) = \frac{14}{50} = \frac{7}{25}
$$

Then, use the identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $ to find:

$$
\sin(\theta) = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

Now compute the area using the formula $ \frac{1}{2}ab\sin(\theta) $:

$$
\text{Area} = \frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{24}{25} = \frac{25}{2} \cdot \frac{24}{25} = \frac{24}{2} = 12
$$

---

### **Conclusion**

All methods—Heron's formula, base-height, coordinate geometry, and trigonometry—consistently yield the same result. Therefore, the area of the triangle is:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of a triangle with sides 5, 5, and 6 units. Hmm, let me think. First, since two sides are equal (both 5 units), this must be an isosceles triangle. I remember there are different ways to calculate the area of a triangle. The most common one is the base times height divided by two, right? But I don't know the height here. Maybe I can figure it out?

Alternatively, there's Heron's formula, which uses the lengths of all three sides. Let me recall how that works. Heron's formula says that if a triangle has sides of length a, b, and c, then the area is the square root of [s(s - a)(s - b)(s - c)], where s is the semi-perimeter. Let me try that.

First, let me calculate the semi-perimeter. The perimeter is 5 + 5 + 6, which is 16. So the semi-perimeter s would be 16 divided by 2, which is 8. Okay, so s = 8.

Now plugging into Heron's formula: Area = sqrt[s(s - a)(s - b)(s - c)]. Let me substitute the values. So that would be sqrt[8*(8 - 5)*(8 - 5)*(8 - 6)]. Let me compute each term:

s - a = 8 - 5 = 3

s - b = 8 - 5 = 3 (since b is also 5)

s - c = 8 - 6 = 2

So multiplying all together: 8 * 3 * 3 * 2. Let me compute that step by step. 8*3 is 24, then 24*3 is 72, then 72*2 is 144. So the area is sqrt(144). The square root of 144 is 12. So the area is 12 square units? Wait, that seems straightforward. Let me check if I did that right.

Alternatively, maybe I can use the base and height method. Let me try that to verify. Let's take the base as 6 units. Then, since it's an isosceles triangle, the height can be found by splitting the base into two equal parts. So the base is 6, so each half is 3 units. Then, the height forms a right triangle with the side of 5 units and the base of 3 units. So using Pythagoras: height squared + 3 squared = 5 squared. So height squared = 25 - 9 = 16. Therefore, height is 4. Then area is (base * height)/2 = (6 * 4)/2 = 24/2 = 12. Okay, same answer. So that confirms it. So the area is 12 square units.

Wait, so both methods give me 12. That seems solid. Let me just think if there's another way to check. Maybe using coordinates? Let me try that.

Let me place the triangle on a coordinate system. Let me put the base of 6 units along the x-axis, from (0, 0) to (6, 0). The third vertex is somewhere above the base. Since it's an isosceles triangle with sides 5, 5, 6, the third vertex should be centered above the base. Wait, the base is 6 units, so the midpoint is at (3, 0). The height we found earlier is 4, so the third vertex is at (3, 4). Then, the three vertices are (0, 0), (6, 0), and (3, 4). Let me compute the area using coordinates. There's the shoelace formula, which is another way.

Shoelace formula: For coordinates (x1, y1), (x2, y2), (x3, y3), the area is |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|.

Plugging in the points:

x1 = 0, y1 = 0

x2 = 6, y2 = 0

x3 = 3, y3 = 4

So compute:

0*(0 - 4) + 6*(4 - 0) + 3*(0 - 0) all over 2, absolute value.

Calculates to:

0 + 6*4 + 3*0 = 24. Then divide by 2: 24/2 = 12. Absolute value is 12. So again, same answer. So that's three different methods giving me 12. That seems pretty confident.

Alternatively, maybe I can use vectors or trigonometry? Let me try trigonometry. If I can find an angle, then use the formula (1/2)*ab*sin(theta). Let me see. Let me consider the two sides of 5 units and the included angle. Wait, if I can find the angle between the two equal sides, then I can compute the area.

But how do I find that angle? Well, using the Law of Cosines. For a triangle with sides a, b, c, c² = a² + b² - 2ab*cos(theta), where theta is the angle between sides a and b.

In this case, let me take the two sides of 5 units and the included angle. The third side is 6 units. So:

6² = 5² + 5² - 2*5*5*cos(theta)

So 36 = 25 + 25 - 50*cos(theta)

Simplify: 36 = 50 - 50*cos(theta)

Subtract 50: 36 - 50 = -50*cos(theta)

-14 = -50*cos(theta)

Divide both sides by -50: cos(theta) = 14/50 = 7/25

So theta = arccos(7/25). Then, the area is (1/2)*ab*sin(theta) = (1/2)*5*5*sin(theta) = (25/2)*sin(theta)

Now, I need to find sin(theta). Since cos(theta) = 7/25, then sin(theta) can be found using sin²(theta) + cos²(theta) = 1. Therefore, sin(theta) = sqrt(1 - (7/25)^2) = sqrt(1 - 49/625) = sqrt(576/625) = 24/25. Since theta is an angle in a triangle, it's between 0 and 180 degrees, so sine is positive.

Therefore, area is (25/2)*(24/25) = (24/2) = 12. Again, same answer. Wow, so all methods lead to 12. That's reassuring.

So, I think the area is definitely 12 square units. I don't see any mistakes in my calculations. All different approaches confirm the same result. So I can be confident that the answer is 12.

**Final Answer**
The area of the triangle is \boxed{12} square units.

To find the area of a triangle with sides measuring 5, 5, and 6 units, we can use multiple approaches. Here's a step-by-step breakdown of the most effective method, followed by verification using alternative techniques.

---

### **Step 1: Use Heron's Formula**

Heron's formula is a reliable method for calculating the area of a triangle when all three side lengths are known. The formula is:

$$
\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}
$$

Where:
- $ a = 5 $, $ b = 5 $, $ c = 6 $
- $ s $ is the semi-perimeter:
  $$
  s = \frac{a + b + c}{2} = \frac{5 + 5 + 6}{2} = 8
  $$

Now plug into the formula:

$$
\text{Area} = \sqrt{8(8 - 5)(8 - 5)(8 - 6)} = \sqrt{8 \cdot 3 \cdot 3 \cdot 2}
$$

$$
= \sqrt{144} = 12
$$

---

### **Step 2: Verify Using Base and Height**

Since the triangle is isosceles (two sides of 5 units), we can drop a perpendicular from the vertex opposite the base (6 units) to the base, splitting it into two equal segments of 3 units each. This forms two right triangles with:

- Hypotenuse = 5 units
- One leg = 3 units
- The other leg = height $ h $

Using the Pythagorean Theorem:

$$
h^2 + 3^2 = 5^2 \Rightarrow h^2 = 25 - 9 = 16 \Rightarrow h = 4
$$

Now compute the area:

$$
\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 4 = 12
$$

---

### **Step 3: Confirm with Coordinate Geometry (Shoelace Formula)**

Place the triangle on the coordinate plane with vertices at:
- $ A = (0, 0) $
- $ B = (6, 0) $
- $ C = (3, 4) $

Using the shoelace formula:

$$
\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|
$$

$$
= \frac{1}{2} |0(0 - 4) + 6(4 - 0) + 3(0 - 0)| = \frac{1}{2} |0 + 24 + 0| = \frac{1}{2} \cdot 24 = 12
$$

---

### **Step 4: Use Trigonometry (Law of Cosines and Sine)**

Using the Law of Cosines to find the angle between the two equal sides:

$$
6^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \cos(\theta)
\Rightarrow 36 = 25 + 25 - 50\cos(\theta)
\Rightarrow \cos(\theta) = \frac{14}{50} = \frac{7}{25}
$$

Then, use the identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $ to find:

$$
\sin(\theta) = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

Now compute the area using the formula $ \frac{1}{2}ab\sin(\theta) $:

$$
\text{Area} = \frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{24}{25} = \frac{25}{2} \cdot \frac{24}{25} = \frac{24}{2} = 12
$$

---

### **Conclusion**

All methods—Heron's formula, base-height, coordinate geometry, and trigonometry—consistently yield the same result. Therefore, the area of the triangle is:

$$
\boxed{12}
$$

Solution: This triangle is isosceles, and so the altitude to the side with length 6 must hit that side at its midpoint. Thus our triangle is divided into two right triangles with hypotenuse $5$ and one side of length $3$. Thus each of these is a $3-4-5$ triangle, and each one has area $\frac{3 \times 4}{2} = 6$, for a total area of $\boxed{12}$.