Explanation:
Okay, so I need to find the degree measure of the least positive angle θ such that tanθ equals this big fraction. Let me write down the problem again to make sure I have it right:

$$
\tan \theta = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}
$$

Alright, so the numerator and denominator are both combinations of sine and cosine terms. Maybe I can simplify them using trigonometric identities? Let me recall some identities. The numerator has terms like cos A cos B and sin A sin B, which remind me of the cosine addition formula. Similarly, the denominator has terms like sin A cos B and cos A sin B, which are similar to the sine addition formula.

Let me recall:

cos(A + B) = cos A cos B - sin A sin B

sin(A + B) = sin A cos B + cos A sin B

Wait, so the numerator has cos5 cos20 - sin5 sin20, which is cos(5 + 20) = cos25°, and then there's also cos35 cos50 - sin35 sin50, which would be cos(35 + 50) = cos85°. So maybe the numerator can be simplified as cos25° + cos85°?

Wait, let me check:

Original numerator:

cos5° cos20° - sin5° sin20° + cos35° cos50° - sin35° sin50°

Yes, that's exactly cos(5° + 20°) + cos(35° + 50°) = cos25° + cos85°. So numerator simplifies to cos25° + cos85°.

Similarly, let me look at the denominator:

sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°

Hmm. Let me group terms:

First, sin5° cos20° + cos5° sin20°, which is sin(5° + 20°) = sin25°, and then - sin35° cos50° - cos35° sin50°, which is - [sin35° cos50° + cos35° sin50°] = - sin(35° + 50°) = - sin85°. Therefore, the denominator simplifies to sin25° - sin85°.

So putting it all together, tanθ = [cos25° + cos85°] / [sin25° - sin85°]. Now, I need to simplify this expression.

Let me recall some sum-to-product identities. For the numerator, cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2], and for the denominator, sin A - sin B = 2 cos[(A+B)/2] sin[(A-B)/2].

Let me apply these identities.

First, numerator: cos25° + cos85°. Let me set A = 25°, B = 85°. Then:

cos25° + cos85° = 2 cos[(25 + 85)/2] cos[(25 - 85)/2] = 2 cos(110°/2) cos(-60°/2) = 2 cos55° cos(-30°). Since cosine is even, cos(-30°) = cos30°, so this becomes 2 cos55° cos30°.

Similarly, denominator: sin25° - sin85°. Let me use the identity sin A - sin B = 2 cos[(A+B)/2] sin[(A - B)/2]. Let me set A = 25°, B = 85°, so:

sin25° - sin85° = 2 cos[(25 + 85)/2] sin[(25 - 85)/2] = 2 cos(110°/2) sin(-60°/2) = 2 cos55° sin(-30°). Since sine is odd, sin(-30°) = -sin30°, so this becomes 2 cos55° (-sin30°) = -2 cos55° sin30°.

Therefore, tanθ = [2 cos55° cos30°] / [-2 cos55° sin30°]. The 2 cos55° cancels out from numerator and denominator, leaving tanθ = [cos30°] / [-sin30°] = -cot30°.

Wait, cot30° is 1/tan30°, which is √3. So tanθ = -√3. Therefore, θ is an angle whose tangent is -√3. But we need the least positive angle θ. Let me recall that tanθ = -√3. The tangent function is negative in the second and fourth quadrants. The reference angle for tanθ = √3 is 60°, since tan60° = √3. Therefore, the angles where tanθ = -√3 are 180° - 60° = 120° and 360° - 60° = 300°, and so on. The least positive angle would be 120°, since 120° is less than 300°.

Wait, but let me check if I did all the steps correctly. Let me go back through the simplification.

Starting with numerator:

cos5° cos20° - sin5° sin20° + cos35° cos50° - sin35° sin50°

Yes, that's cos(5+20) + cos(35+50) = cos25 + cos85. Then using sum to product:

cos25 + cos85 = 2 cos[(25+85)/2] cos[(25-85)/2] = 2 cos(55) cos(-30) = 2 cos55 cos30. Correct.

Denominator:

sin5 cos20 + cos5 sin20 - sin35 cos50 - cos35 sin50

Which is sin(5+20) - sin(35+50) = sin25 - sin85. Then using identity:

sin25 - sin85 = 2 cos[(25+85)/2] sin[(25-85)/2] = 2 cos55 sin(-30) = -2 cos55 sin30. Correct.

Therefore tanθ = [2 cos55 cos30]/[-2 cos55 sin30] = -cot30 = -√3. Correct.

So tanθ = -√3. So θ is in which quadrant? Since tan is negative, θ is in second or fourth quadrant. The principal value for arctangent of -√3 would be -60°, but since we need the least positive angle, we add 180° to -60°, getting 120°, which is in the second quadrant. Alternatively, 360° - 60° = 300°, but 120° is smaller. So θ = 120°. Therefore, the answer is 120 degrees.

But let me check if there's a mistake here. Let me verify with actual calculation. Let me compute numerator and denominator numerically to check.

First, compute numerator:

cos5° cos20° + cos35° cos50° - sin5° sin20° - sin35° sin50°

Let me compute each term:

cos5° ≈ 0.9962, cos20° ≈ 0.9397, so cos5° cos20° ≈ 0.9962 * 0.9397 ≈ 0.935.

cos35° ≈ 0.8192, cos50° ≈ 0.6428, so cos35° cos50° ≈ 0.8192 * 0.6428 ≈ 0.527.

sin5° ≈ 0.0872, sin20° ≈ 0.3420, so sin5° sin20° ≈ 0.0872 * 0.3420 ≈ 0.0298.

sin35° ≈ 0.5736, sin50° ≈ 0.7660, so sin35° sin50° ≈ 0.5736 * 0.7660 ≈ 0.439.

Therefore numerator ≈ 0.935 + 0.527 - 0.0298 - 0.439 ≈ 0.935 + 0.527 = 1.462; 1.462 - 0.0298 = 1.4322; 1.4322 - 0.439 ≈ 0.9932.

Now denominator:

sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°

Compute each term:

sin5° cos20° ≈ 0.0872 * 0.9397 ≈ 0.0819.

-sin35° cos50° ≈ -0.5736 * 0.6428 ≈ -0.368.

cos5° sin20° ≈ 0.9962 * 0.3420 ≈ 0.3407.

-cos35° sin50° ≈ -0.8192 * 0.7660 ≈ -0.628.

Adding them up: 0.0819 - 0.368 + 0.3407 - 0.628 ≈

First, 0.0819 - 0.368 ≈ -0.2861

Then, -0.2861 + 0.3407 ≈ 0.0546

Then, 0.0546 - 0.628 ≈ -0.5734

So denominator ≈ -0.5734

Therefore tanθ ≈ 0.9932 / (-0.5734) ≈ -1.731

Now tan60° is √3 ≈ 1.732, so tanθ ≈ -1.731 is approximately -√3, which matches our previous result. Therefore tanθ ≈ -√3, so θ ≈ 120°, since tan120° = tan(180° - 60°) = -tan60° = -√3. So that checks out.

Therefore, the least positive angle θ is 120 degrees. Therefore, the answer is 120°, so boxed answer is \boxed{120}.

But let me just check if there's another angle smaller than 120° that could satisfy tanθ = -√3. Since tanθ is periodic with period 180°, the solutions are θ = -60° + 180°k, where k is integer. The least positive angle would be when k=1: -60° + 180° = 120°, and k=0 gives -60°, which is negative. So yes, 120° is indeed the least positive angle. Therefore, the answer is 120°.

**Final Answer**
\boxed{120}

To solve the given problem, we are asked to find the **least positive angle** $ \theta $ such that:

$$
\tan \theta = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}
$$

---

### Step 1: Simplify the Numerator

We observe that the numerator contains terms of the form $ \cos A \cos B - \sin A \sin B $, which is the **cosine addition formula**:

$$
\cos(A + B) = \cos A \cos B - \sin A \sin B
$$

So, we can group the terms:

$$
\cos 5^\circ \cos 20^\circ - \sin 5^\circ \sin 20^\circ = \cos(5^\circ + 20^\circ) = \cos 25^\circ
$$
$$
\cos 35^\circ \cos 50^\circ - \sin 35^\circ \sin 50^\circ = \cos(35^\circ + 50^\circ) = \cos 85^\circ
$$

Thus, the **numerator** becomes:

$$
\cos 25^\circ + \cos 85^\circ
$$

---

### Step 2: Simplify the Denominator

The denominator contains terms of the form $ \sin A \cos B + \cos A \sin B $, which is the **sine addition formula**:

$$
\sin(A + B) = \sin A \cos B + \cos A \sin B
$$

So, we can group the terms:

$$
\sin 5^\circ \cos 20^\circ + \cos 5^\circ \sin 20^\circ = \sin(5^\circ + 20^\circ) = \sin 25^\circ
$$
$$
-\left( \sin 35^\circ \cos 50^\circ + \cos 35^\circ \sin 50^\circ \right) = -\sin(35^\circ + 50^\circ) = -\sin 85^\circ
$$

Thus, the **denominator** becomes:

$$
\sin 25^\circ - \sin 85^\circ
$$

---

### Step 3: Apply Sum-to-Product Identities

We now simplify both the numerator and denominator using **sum-to-product identities**:

#### Numerator:
$$
\cos 25^\circ + \cos 85^\circ = 2 \cos\left( \frac{25^\circ + 85^\circ}{2} \right) \cos\left( \frac{25^\circ - 85^\circ}{2} \right)
= 2 \cos(55^\circ) \cos(-30^\circ)
$$

Since $ \cos(-x) = \cos x $, this becomes:

$$
2 \cos(55^\circ) \cos(30^\circ)
$$

#### Denominator:
$$
\sin 25^\circ - \sin 85^\circ = 2 \cos\left( \frac{25^\circ + 85^\circ}{2} \right) \sin\left( \frac{25^\circ - 85^\circ}{2} \right)
= 2 \cos(55^\circ) \sin(-30^\circ)
$$

Since $ \sin(-x) = -\sin x $, this becomes:

$$
-2 \cos(55^\circ) \sin(30^\circ)
$$

---

### Step 4: Compute the Tangent

Now, we compute:

$$
\tan \theta = \frac{2 \cos(55^\circ) \cos(30^\circ)}{-2 \cos(55^\circ) \sin(30^\circ)} = \frac{\cos(30^\circ)}{-\sin(30^\circ)} = -\cot(30^\circ)
$$

Recall that $ \cot(30^\circ) = \frac{1}{\tan(30^\circ)} = \sqrt{3} $, so:

$$
\tan \theta = -\sqrt{3}
$$

---

### Step 5: Find the Least Positive Angle

We are looking for the **least positive angle** $ \theta $ such that $ \tan \theta = -\sqrt{3} $.

The tangent function is negative in the **second and fourth quadrants**. The reference angle for $ \tan^{-1}(\sqrt{3}) $ is $ 60^\circ $, so the angles where $ \tan \theta = -\sqrt{3} $ are:

- $ \theta = 180^\circ - 60^\circ = 120^\circ $
- $ \theta = 360^\circ - 60^\circ = 300^\circ $

The **least positive** such angle is:

$$
\boxed{120}
$$

---

### Final Answer

$$
\boxed{120}
$$

[Assistant]
Explanation:
Okay, so I need to find the degree measure of the least positive angle θ such that tanθ equals this big fraction. Let me write down the problem again to make sure I have it right:

$$
\tan \theta = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}
$$

Alright, so the numerator and denominator are both combinations of sine and cosine terms. Maybe I can simplify them using trigonometric identities? Let me recall some identities. The numerator has terms like cos A cos B and sin A sin B, which remind me of the cosine addition formula. Similarly, the denominator has terms like sin A cos B and cos A sin B, which are similar to the sine addition formula.

Let me recall:

cos(A + B) = cos A cos B - sin A sin B

sin(A + B) = sin A cos B + cos A sin B

Wait, so the numerator has cos5 cos20 - sin5 sin20, which is cos(5 + 20) = cos25°, and then there's also cos35 cos50 - sin35 sin50, which would be cos(35 + 50) = cos85°. So maybe the numerator can be simplified as cos25° + cos85°?

Wait, let me check:

Original numerator:

cos5° cos20° - sin5° sin20° + cos35° cos50° - sin35° sin50°

Yes, that's exactly cos(5° + 20°) + cos(35° + 50°) = cos25° + cos85°. So numerator simplifies to cos25° + cos85°.

Similarly, let me look at the denominator:

sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°

Hmm. Let me group terms:

First, sin5° cos20° + cos5° sin20°, which is sin(5° + 20°) = sin25°, and then - sin35° cos50° - cos35° sin50°, which is - [sin35° cos50° + cos35° sin50°] = - sin(35° + 50°) = - sin85°. Therefore, the denominator simplifies to sin25° - sin85°.

So putting it all together, tanθ = [cos25° + cos85°] / [sin25° - sin85°]. Now, I need to simplify this expression.

Let me recall some sum-to-product identities. For the numerator, cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2], and for the denominator, sin A - sin B = 2 cos[(A+B)/2] sin[(A-B)/2].

Let me apply these identities.

First, numerator: cos25° + cos85°. Let me set A = 25°, B = 85°. Then:

cos25° + cos85° = 2 cos[(25 + 85)/2] cos[(25 - 85)/2] = 2 cos(110°/2) cos(-60°/2) = 2 cos55° cos(-30°). Since cosine is even, cos(-30°) = cos30°, so this becomes 2 cos55° cos30°.

Similarly, denominator: sin25° - sin85°. Let me use the identity sin A - sin B = 2 cos[(A+B)/2] sin[(A - B)/2]. Let me set A = 25°, B = 85°, so:

sin25° - sin85° = 2 cos[(25 + 85)/2] sin[(25 - 85)/2] = 2 cos(110°/2) sin(-60°/2) = 2 cos55° sin(-30°). Since sine is odd, sin(-30°) = -sin30°, so this becomes 2 cos55° (-sin30°) = -2 cos55° sin30°.

Therefore, tanθ = [2 cos55° cos30°] / [-2 cos55° sin30°]. The 2 cos55° cancels out from numerator and denominator, leaving tanθ = [cos30°] / [-sin30°] = -cot30°.

Wait, cot30° is 1/tan30°, which is √3. So tanθ = -√3. Therefore, θ is an angle whose tangent is -√3. But we need the least positive angle θ. Let me recall that tanθ = -√3. The tangent function is negative in the second and fourth quadrants. The reference angle for tanθ = √3 is 60°, since tan60° = √3. Therefore, the angles where tanθ = -√3 are 180° - 60° = 120° and 360° - 60° = 300°, and so on. The least positive angle would be 120°, since 120° is less than 300°.

Wait, but let me check if I did all the steps correctly. Let me go back through the simplification.

Starting with numerator:

cos5° cos20° - sin5° sin20° + cos35° cos50° - sin35° sin50°

Yes, that's cos(5+20) + cos(35+50) = cos25 + cos85. Then using sum to product:

cos25 + cos85 = 2 cos[(25+85)/2] cos[(25-85)/2] = 2 cos(55) cos(-30) = 2 cos55 cos30. Correct.

Denominator:

sin5 cos20 + cos5 sin20 - sin35 cos50 - cos35 sin50

Which is sin(5+20) - sin(35+50) = sin25 - sin85. Then using identity:

sin25 - sin85 = 2 cos[(25+85)/2] sin[(25-85)/2] = 2 cos55 sin(-30) = -2 cos55 sin30. Correct.

Therefore tanθ = [2 cos55 cos30]/[-2 cos55 sin30] = -cot30 = -√3. Correct.

So tanθ = -√3. So θ is in which quadrant? Since tan is negative, θ is in second or fourth quadrant. The principal value for arctangent of -√3 would be -60°, but since we need the least positive angle, we add 180° to -60°, getting 120°, which is in the second quadrant. Alternatively, 360° - 60° = 300°, but 120° is smaller. So θ = 120°. Therefore, the answer is 120 degrees.

But let me check if there's a mistake here. Let me verify with actual calculation. Let me compute numerator and denominator numerically to check.

First, compute numerator:

cos5° cos20° + cos35° cos50° - sin5° sin20° - sin35° sin50°

Let me compute each term:

cos5° ≈ 0.9962, cos20° ≈ 0.9397, so cos5° cos20° ≈ 0.9962 * 0.9397 ≈ 0.935.

cos35° ≈ 0.8192, cos50° ≈ 0.6428, so cos35° cos50° ≈ 0.8192 * 0.6428 ≈ 0.527.

sin5° ≈ 0.0872, sin20° ≈ 0.3420, so sin5° sin20° ≈ 0.0872 * 0.3420 ≈ 0.0298.

sin35° ≈ 0.5736, sin50° ≈ 0.7660, so sin35° sin50° ≈ 0.5736 * 0.7660 ≈ 0.439.

Therefore numerator ≈ 0.935 + 0.527 - 0.0298 - 0.439 ≈ 0.935 + 0.527 = 1.462; 1.462 - 0.0298 = 1.4322; 1.4322 - 0.439 ≈ 0.9932.

Now denominator:

sin5° cos20° - sin35° cos50° + cos5° sin20° - cos35° sin50°

Compute each term:

sin5° cos20° ≈ 0.0872 * 0.9397 ≈ 0.0819.

-sin35° cos50° ≈ -0.5736 * 0.6428 ≈ -0.368.

cos5° sin20° ≈ 0.9962 * 0.3420 ≈ 0.3407.

-cos35° sin50° ≈ -0.8192 * 0.7660 ≈ -0.628.

Adding them up: 0.0819 - 0.368 + 0.3407 - 0.628 ≈

First, 0.0819 - 0.368 ≈ -0.2861

Then, -0.2861 + 0.3407 ≈ 0.0546

Then, 0.0546 - 0.628 ≈ -0.5734

So denominator ≈ -0.5734

Therefore tanθ ≈ 0.9932 / (-0.5734) ≈ -1.731

Now tan60° is √3 ≈ 1.732, so tanθ ≈ -1.731 is approximately -√3, which matches our previous result. Therefore tanθ ≈ -√3, so θ ≈ 120°, since tan120° = tan(180° - 60°) = -tan60° = -√3. So that checks out.

Therefore, the least positive angle θ is 120 degrees. Therefore, the answer is 120°, so boxed answer is \boxed{120}.

But let me just check if there's another angle smaller than 120° that could satisfy tanθ = -√3. Since tanθ is periodic with period 180°, the solutions are θ = -60° + 180°k, where k is integer. The least positive angle would be when k=1: -60° + 180° = 120°, and k=0 gives -60°, which is negative. So yes, 120° is indeed the least positive angle. Therefore, the answer is 120°.

**Final Answer**
\boxed{120}

To solve the given problem, we are asked to find the **least positive angle** $ \theta $ such that:

$$
\tan \theta = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}
$$

---

### Step 1: Simplify the Numerator

We observe that the numerator contains terms of the form $ \cos A \cos B - \sin A \sin B $, which is the **cosine addition formula**:

$$
\cos(A + B) = \cos A \cos B - \sin A \sin B
$$

So, we can group the terms:

$$
\cos 5^\circ \cos 20^\circ - \sin 5^\circ \sin 20^\circ = \cos(5^\circ + 20^\circ) = \cos 25^\circ
$$
$$
\cos 35^\circ \cos 50^\circ - \sin 35^\circ \sin 50^\circ = \cos(35^\circ + 50^\circ) = \cos 85^\circ
$$

Thus, the **numerator** becomes:

$$
\cos 25^\circ + \cos 85^\circ
$$

---

### Step 2: Simplify the Denominator

The denominator contains terms of the form $ \sin A \cos B + \cos A \sin B $, which is the **sine addition formula**:

$$
\sin(A + B) = \sin A \cos B + \cos A \sin B
$$

So, we can group the terms:

$$
\sin 5^\circ \cos 20^\circ + \cos 5^\circ \sin 20^\circ = \sin(5^\circ + 20^\circ) = \sin 25^\circ
$$
$$
-\left( \sin 35^\circ \cos 50^\circ + \cos 35^\circ \sin 50^\circ \right) = -\sin(35^\circ + 50^\circ) = -\sin 85^\circ
$$

Thus, the **denominator** becomes:

$$
\sin 25^\circ - \sin 85^\circ
$$

---

### Step 3: Apply Sum-to-Product Identities

We now simplify both the numerator and denominator using **sum-to-product identities**:

#### Numerator:
$$
\cos 25^\circ + \cos 85^\circ = 2 \cos\left( \frac{25^\circ + 85^\circ}{2} \right) \cos\left( \frac{25^\circ - 85^\circ}{2} \right)
= 2 \cos(55^\circ) \cos(-30^\circ)
$$

Since $ \cos(-x) = \cos x $, this becomes:

$$
2 \cos(55^\circ) \cos(30^\circ)
$$

#### Denominator:
$$
\sin 25^\circ - \sin 85^\circ = 2 \cos\left( \frac{25^\circ + 85^\circ}{2} \right) \sin\left( \frac{25^\circ - 85^\circ}{2} \right)
= 2 \cos(55^\circ) \sin(-30^\circ)
$$

Since $ \sin(-x) = -\sin x $, this becomes:

$$
-2 \cos(55^\circ) \sin(30^\circ)
$$

---

### Step 4: Compute the Tangent

Now, we compute:

$$
\tan \theta = \frac{2 \cos(55^\circ) \cos(30^\circ)}{-2 \cos(55^\circ) \sin(30^\circ)} = \frac{\cos(30^\circ)}{-\sin(30^\circ)} = -\cot(30^\circ)
$$

Recall that $ \cot(30^\circ) = \frac{1}{\tan(30^\circ)} = \sqrt{3} $, so:

$$
\tan \theta = -\sqrt{3}
$$

---

### Step 5: Find the Least Positive Angle

We are looking for the **least positive angle** $ \theta $ such that $ \tan \theta = -\sqrt{3} $.

The tangent function is negative in the **second and fourth quadrants**. The reference angle for $ \tan^{-1}(\sqrt{3}) $ is $ 60^\circ $, so the angles where $ \tan \theta = -\sqrt{3} $ are:

- $ \theta = 180^\circ - 60^\circ = 120^\circ $
- $ \theta = 360^\circ - 60^\circ = 300^\circ $

The **least positive** such angle is:

$$
\boxed{120}
$$

---

### Final Answer

$$
\boxed{120}
$$

Solution:
From the angle addition formula, the numerator is
\begin{align*}
&(\cos 5^\circ \cos 20^\circ - \sin 5^\circ \sin 20^\circ) + (\cos 35^\circ \cos 50^\circ -  \sin 35^\circ \sin 50^\circ) \\
&= \cos (5^\circ + 20^\circ) + \cos (35^\circ + 50^\circ) \\
&= \cos 25^\circ + \cos 85^\circ.
\end{align*}From the sum-to-product formula, $\cos 25^\circ + \cos 85^\circ = 2 \cos 55^\circ \cos 30^\circ.$

Similarly, the denominator is
\begin{align*}
&\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ) \\
&= (\sin 5^\circ \cos 20^\circ +  \cos 5^\circ \sin 20^\circ) - (\sin 35^\circ \cos 50^\circ + \cos 35^\circ \sin 50^\circ) \\
&= \sin (5^\circ + 20^\circ) - \sin (35^\circ + 50^\circ) \\
&= \sin 25^\circ - \sin 85^\circ \\
&= -2 \sin 30^\circ \cos 55^\circ,
\end{align*}so the expression is equal to
\[\frac{2 \cos 55^\circ \cos 30^\circ}{-2 \sin 30^\circ \cos 55^\circ} = -\frac{\cos 30^\circ}{\sin 30^\circ} = -\sqrt{3} = \tan 120^\circ.\]Hence, the smallest such $\theta$ is $\boxed{120^\circ}.$