Explanation:
Okay, so I need to find the measure of angle AHB in triangle ABC where AD and BE are altitudes intersecting at H. They gave me angles BAC and ABC as 54 degrees and 52 degrees respectively. Let me try to visualize the triangle first.

First, let me sketch triangle ABC. Let me recall that in any triangle, the sum of the angles is 180 degrees. So, if angle BAC is 54° and angle ABC is 52°, then angle ACB can be calculated. Let me do that first.

So, angle BAC + angle ABC + angle ACB = 180°
54° + 52° + angle ACB = 180°
Adding 54 and 52 gives 106°, so angle ACB = 180° - 106° = 74°. Okay, so angle C is 74 degrees.

Now, AD and BE are altitudes. Let me recall that an altitude of a triangle is a perpendicular segment from a vertex to the opposite side (or the line containing the opposite side). So, AD is the altitude from A to BC, meaning AD is perpendicular to BC. Similarly, BE is the altitude from B to AC, so BE is perpendicular to AC.

They intersect at H, which is the orthocenter of triangle ABC. The orthocenter is the point where all three altitudes intersect. So, H is the orthocenter here.

Now, the question is asking for angle AHB. So, angle at H between points A and B. Let me think about how to find this angle.

I remember that in a triangle, the measure of the angle formed by two altitudes can be related to the angles of the triangle. Maybe there's a formula or a relationship here. Alternatively, maybe I can use some properties of cyclic quadrilaterals or triangle angle sums.

Alternatively, maybe I can find coordinates or use trigonometry. But since angles are given, perhaps there's a more straightforward geometric approach.

Let me recall that in triangle ABC, the orthocenter H creates several right angles with the sides. For example, since AD is an altitude, angle ADB is 90°, and similarly, angle BEA is 90°. Maybe I can use these right angles to find some relationships.

Alternatively, maybe I can consider quadrilateral HDCE or something like that? Wait, not sure. Let me think.

Alternatively, since H is the orthocenter, maybe I can find some angles in triangle AHB. Let me consider triangle AHB. If I can find some angles in this triangle, then I can compute angle AHB.

Let me note that in triangle ABC, angles at A, B, C are 54°, 52°, 74° respectively. Let me try to find some relationships involving H.

Since AD and BE are altitudes, then AD ⊥ BC and BE ⊥ AC. Let me consider angles involving H.

Let me think about the angles at H. For example, angle AHB. Let me recall that in some cases, the measure of angle AHB can be related to the angles of the original triangle. For example, in an acute triangle, the orthocenter lies inside the triangle, and angle AHB is equal to 180° minus angle ACB. Wait, is that true?

Wait, let me check. Let me recall that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is that a general formula?

Wait, let me think. Let me consider quadrilateral HDCE, but maybe not. Alternatively, let me consider triangle AHB. Let me try to find angles at H.

Alternatively, since AD and BE are altitudes, then angles at D and E are right angles. Let me consider triangle AHB. Let me try to find angles at A and B in triangle AHB.

Wait, angle at A in triangle AHB is the same as angle between HA and AB. Wait, but HA is part of the altitude AD. Wait, maybe not. Let me think.

Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several similar triangles or something.

Alternatively, maybe I can use the fact that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Wait, if that's the case, then angle AHB would be 180° - 74° = 106°. But I need to verify if this is correct.

Wait, let me think again. Let me consider triangle AHB. Let me try to find angles at H. Let me think about the angles formed by the altitudes.

Since AD is an altitude, angle ADB is 90°, and BE is an altitude, angle BEA is 90°. Let me consider quadrilateral AEHD. Wait, maybe not. Let me think about triangle AHB.

Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates angles that are supplementary to the original angles. Wait, here's a thought: In triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Let me check this with an example.

Suppose we have an equilateral triangle, all angles 60°, then the orthocenter coincides with the centroid and the circumcenter. Then angle AHB would be... Well, in that case, all angles at H would be 120°, since each angle between two altitudes would be 120°, which is 180° - 60°, so that seems to fit. So maybe that formula is correct.

So if angle ACB is 74°, then angle AHB would be 180° - 74° = 106°. But let me verify this with another approach to be sure.

Alternatively, let me try to compute angle AHB using other methods.

Let me consider triangle AHB. Let me find angles at A and B in triangle AHB. Wait, angle at A in triangle AHB is the same as angle between HA and AB. But HA is part of the altitude from A to BC. Let me think.

Alternatively, since AD is an altitude, then AD is perpendicular to BC. Similarly, BE is perpendicular to AC. Let me consider angles involving H.

Let me consider triangle AHB. Let me denote angle HAB and angle HBA, then angle AHB = 180° - angle HAB - angle HBA.

So, if I can find angle HAB and angle HBA, then I can compute angle AHB.

Let me find angle HAB. Since AD is an altitude, angle BAD is equal to 90° - angle ABC. Wait, let me think.

In triangle ABC, angle at B is 52°, and AD is the altitude from A to BC. So, in triangle ABD, which is a right triangle (since AD is an altitude), angle at D is 90°, angle at B is 52°, so angle BAD = 90° - angle ABC = 90° - 52° = 38°. Therefore, angle BAD is 38°, which is angle between BA and AD. But angle HAB is the same as angle BAD because H lies on AD. Therefore, angle HAB is 38°.

Similarly, let me find angle HBA. Since BE is the altitude from B to AC, so in triangle BEA, which is a right triangle, angle at E is 90°, angle at A is 54°, so angle ABE = 90° - angle BAC = 90° - 54° = 36°. Therefore, angle ABE is 36°, which is the angle between BA and BE. Since H lies on BE, angle HBA is equal to angle ABE, which is 36°.

Therefore, in triangle AHB, angles at A and B are 38° and 36°, so angle AHB = 180° - 38° - 36° = 180° - 74° = 106°. So that gives angle AHB as 106°, which matches the previous thought.

Therefore, the answer is 106 degrees. Let me check again if this is correct.

Wait, let me verify the steps again. First, angle BAD is 90° - angle ABC. Is that correct?

Yes, because in triangle ABD, right-angled at D, angle at B is angle ABC = 52°, so angle BAD = 90° - 52° = 38°, correct. Similarly, angle ABE is 90° - angle BAC = 90° - 54° = 36°, correct. Then, in triangle AHB, angles at A and B are 38° and 36°, so angle AHB is 180 - 38 - 36 = 106°, which is 180 - 74°, which is 180 - angle ACB. So that seems to hold.

Alternatively, another way to think about this is that angle AHB = 180° - angle ABC - angle BAC? Wait, no, that would be 180 - 54 - 52 = 74°, which is angle ACB. But that's not the case here. Wait, but angle AHB is 106°, which is 180 - 74°, so angle AHB = 180° - angle ACB. So that seems to be a general formula.

So, in general, in any triangle, the measure of angle AHB is equal to 180° minus the measure of angle ACB. Is that a general rule?

Let me think. Let me consider another example. Suppose we have a triangle where angle C is 90°, so it's a right-angled triangle. Then the orthocenter H is at point C, because in a right-angled triangle, the altitudes are the legs themselves. So, angle AHB would be angle ACB, which is 90°, but according to the formula 180° - angle ACB, that would be 90°, which matches. Wait, but if H is at C, then angle AHB is angle ACB? Wait, if H is at C, then angle AHB is angle ACB? Wait, no. If H is at C, then points A, H, B would be A, C, B, so angle AHB is angle ACB, which is 90°, and according to the formula 180° - angle ACB would be 90°, which matches. So that works.

Another example: take an equilateral triangle, all angles 60°, then the orthocenter is at the center. Then angle AHB would be 120°, which is 180° - 60°, so that works. So seems like the formula angle AHB = 180° - angle ACB is correct.

Therefore, in this problem, since angle ACB is 74°, angle AHB is 180° - 74° = 106°, so the answer is 106 degrees.

But let me check again with the previous calculation where I found angles HAB and HBA as 38° and 36°, adding up to 74°, so angle AHB is 106°, which is consistent. Therefore, seems correct.

Alternatively, maybe I can use coordinates to verify. Let me try that.

Let me assign coordinates to triangle ABC. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Wait, but maybe it's easier to use trigonometry.

Alternatively, let me use the Law of Sines to find some relations. Wait, but maybe coordinates would be more straightforward.

Let me try coordinates. Let me assume coordinates for triangle ABC. Let me set point A at (0, 0), point B at (b, 0), and point C at coordinates (d, e). But maybe this is too vague. Alternatively, let me construct triangle ABC with given angles.

Alternatively, since angles are given, maybe I can assign lengths based on the Law of Sines.

Let me denote sides as follows: side opposite angle A is BC = a, side opposite angle B is AC = b, side opposite angle C is AB = c.

By Law of Sines:

a / sin A = b / sin B = c / sin C

Given angles:

Angle A = 54°, angle B = 52°, angle C = 74°

So, let me assign some lengths. Let me assume that the circumradius is 1, but maybe it's easier to assign specific lengths.

Alternatively, let me take AB = c, BC = a, AC = b.

But maybe coordinates. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Let me compute coordinates based on angles.

Alternatively, since I need altitudes, maybe coordinates would complicate. Alternatively, maybe use vectors or trigonometry.

Alternatively, since I already have two methods giving me 106°, maybe I can be confident. But just to be thorough, let me try another approach.

Let me recall that in triangle ABC, the orthocenter H. Then, angles involving H can be related to the original angles. For example, in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. This is a known result. Let me try to recall why that is.

Consider the orthocenter H. Then, quadrilateral HECD, where E and D are the feet of the altitudes, is cyclic? Wait, not sure. Alternatively, since AD and BE are altitudes, intersecting at H, then angles at H can be related.

Alternatively, consider that in triangle ABC, angle HAB = 90° - angle ABC, and angle HBA = 90° - angle BAC, as we found earlier. Therefore, in triangle AHB, angles at A and B are 90° - angle ABC and 90° - angle BAC, respectively. Therefore, angle AHB = 180° - [ (90° - angle ABC) + (90° - angle BAC) ] = 180° - 180° + angle ABC + angle BAC = angle ABC + angle BAC.

Wait, hold on! Wait, that would be angle AHB = angle ABC + angle BAC. But angle ABC + angle BAC is 52° + 54° = 106°, which matches our previous result. Wait, but earlier I thought angle AHB = 180° - angle ACB, which is also 180° - 74° = 106°, so both formulas are consistent. Because angle ABC + angle BAC = 180° - angle ACB. Since angle ABC + angle BAC + angle ACB = 180°, so angle ABC + angle BAC = 180° - angle ACB. Therefore, angle AHB = angle ABC + angle BAC = 180° - angle ACB. Therefore, both formulas are equivalent.

Therefore, angle AHB is equal to the sum of angles at A and B, which is 54° + 52° = 106°, or 180° - angle C = 180° - 74° = 106°. So, both ways, same answer.

Therefore, the measure of angle AHB is 106 degrees.

But let me check once more with another method. Let me use coordinates.

Let me try to assign coordinates to triangle ABC. Let me place point A at (0, 0), point B at (1, 0), and point C somewhere in the plane. Let me compute coordinates of C based on angles.

But maybe it's easier to use Law of Sines. Let me assume AB = c, BC = a, AC = b.

From Law of Sines:

a / sin A = b / sin B = c / sin C

Let me assign AB = c = 1 (for simplicity). Then:

a / sin 54° = 1 / sin 74°, so a = sin 54° / sin 74°

Similarly, b / sin 52° = 1 / sin 74°, so b = sin 52° / sin 74°

Let me compute these values numerically.

First, compute sin 54°, sin 52°, sin 74°.

Using calculator:

sin 54° ≈ 0.8090

sin 52° ≈ 0.7880

sin 74° ≈ 0.9613

Therefore:

a ≈ 0.8090 / 0.9613 ≈ 0.8416

b ≈ 0.7880 / 0.9613 ≈ 0.8197

So, sides:

AB = 1

BC ≈ 0.8416

AC ≈ 0.8197

Now, let me assign coordinates. Let me place point A at (0, 0), point B at (1, 0). Now, need to find coordinates of point C.

Let me denote coordinates of C as (x, y). Then, distance from A to C is b ≈ 0.8197, so:

x² + y² = b² ≈ (0.8197)² ≈ 0.6719

Distance from B to C is a ≈ 0.8416, so:

(x - 1)² + y² = a² ≈ (0.8416)² ≈ 0.7083

Subtracting first equation from the second:

(x - 1)² + y² - x² - y² = 0.7083 - 0.6719

Expanding (x - 1)² - x²:

x² - 2x + 1 - x² = -2x + 1 = 0.7083 - 0.6719 ≈ 0.0364

Therefore:

-2x + 1 ≈ 0.0364 => -2x ≈ -0.9636 => x ≈ 0.4818

Then, from x² + y² ≈ 0.6719:

(0.4818)² + y² ≈ 0.6719 => 0.2321 + y² ≈ 0.6719 => y² ≈ 0.4398 => y ≈ √0.4398 ≈ 0.6632

Therefore, coordinates of C are approximately (0.4818, 0.6632)

Now, let me find equations of altitudes AD and BE, then find their intersection H, then compute angle AHB.

First, find altitude AD: from A(0,0) perpendicular to BC.

First, find slope of BC. Coordinates of B(1, 0) and C(0.4818, 0.6632). The slope of BC is (0.6632 - 0)/(0.4818 - 1) = 0.6632 / (-0.5182) ≈ -1.279

Therefore, the slope of BC is approximately -1.279, so the slope of altitude AD, which is perpendicular to BC, is the negative reciprocal: 1 / 1.279 ≈ 0.781

Therefore, equation of altitude AD: passes through A(0,0), slope ≈ 0.781, so equation is y = 0.781x

Now, find equation of altitude BE: from B(1, 0) perpendicular to AC.

First, find slope of AC. Coordinates of A(0,0) and C(0.4818, 0.6632). Slope of AC is (0.6632 - 0)/(0.4818 - 0) ≈ 0.6632 / 0.4818 ≈ 1.376

Therefore, slope of AC ≈ 1.376, so slope of altitude BE, which is perpendicular, is -1 / 1.376 ≈ -0.727

Equation of altitude BE: passes through B(1, 0), slope ≈ -0.727. So equation is y - 0 = -0.727(x - 1), so y = -0.727x + 0.727

Now, find intersection point H of AD and BE.

AD: y = 0.781x

BE: y = -0.727x + 0.727

Set equal:

0.781x = -0.727x + 0.727

0.781x + 0.727x = 0.727

(0.781 + 0.727)x = 0.727

1.508x ≈ 0.727 => x ≈ 0.727 / 1.508 ≈ 0.482

Then y ≈ 0.781 * 0.482 ≈ 0.376

Therefore, coordinates of H are approximately (0.482, 0.376)

Now, need to compute angle AHB. Points A(0,0), H(0.482, 0.376), B(1, 0)

Compute vectors HA and HB.

Vector HA = A - H = (0 - 0.482, 0 - 0.376) = (-0.482, -0.376)

Vector HB = B - H = (1 - 0.482, 0 - 0.376) = (0.518, -0.376)

Now, angle between vectors HA and HB is angle AHB.

Use the dot product formula:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB = (-0.482)(0.518) + (-0.376)(-0.376) ≈ (-0.249) + (0.141) ≈ -0.108

Compute |HA| = sqrt((-0.482)^2 + (-0.376)^2) ≈ sqrt(0.232 + 0.141) ≈ sqrt(0.373) ≈ 0.611

Compute |HB| = sqrt((0.518)^2 + (-0.376)^2) ≈ sqrt(0.268 + 0.141) ≈ sqrt(0.409) ≈ 0.640

Therefore, cos(theta) ≈ -0.108 / (0.611 * 0.640) ≈ -0.108 / 0.391 ≈ -0.276

Therefore, theta ≈ arccos(-0.276) ≈ 106 degrees (since cos(106°) ≈ cos(90° + 16°) = -sin(16°) ≈ -0.2756). Yes! That's very close. So arccos(-0.276) is approximately 106 degrees. Therefore, angle AHB is approximately 106°, which matches our previous results.

Therefore, this coordinate method also gives 106°, so I can be confident that the answer is 106 degrees.

Therefore, the measure of angle AHB is 106 degrees.

**Final Answer**
\boxed{106}

To determine the measure of angle $ \angle AHB $ in triangle $ \triangle ABC $, where altitudes $ \overline{AD} $ and $ \overline{BE} $ intersect at the orthocenter $ H $, we are given the following:

- $ \angle BAC = 54^\circ $
- $ \angle ABC = 52^\circ $

---

### Step 1: Determine the Third Angle of Triangle $ ABC $

In any triangle, the sum of the interior angles is $ 180^\circ $. Therefore, we can compute the third angle:

$$
\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 54^\circ - 52^\circ = 74^\circ
$$

---

### Step 2: Use a Known Geometric Property

A key geometric property of the orthocenter $ H $ in triangle $ ABC $ is that the angle formed at the orthocenter between two vertices (e.g., $ \angle AHB $) is related to the third angle of the triangle. Specifically:

$$
\angle AHB = 180^\circ - \angle ACB
$$

This is because the orthocenter lies inside the triangle (since all angles are acute), and the angle $ \angle AHB $ is formed by two altitudes intersecting at $ H $, which creates a supplementary angle with $ \angle ACB $.

---

### Step 3: Apply the Formula

Using the known value of $ \angle ACB = 74^\circ $, we compute:

$$
\angle AHB = 180^\circ - 74^\circ = 106^\circ
$$

---

### Step 4: Verification via Triangle Angle Sum

Alternatively, we can verify this by considering triangle $ \triangle AHB $. The angles at $ A $ and $ B $ in this triangle are formed by the altitudes:

- $ \angle HAB = 90^\circ - \angle ABC = 90^\circ - 52^\circ = 38^\circ $
- $ \angle HBA = 90^\circ - \angle BAC = 90^\circ - 54^\circ = 36^\circ $

Then, in triangle $ \triangle AHB $, the angle at $ H $ is:

$$
\angle AHB = 180^\circ - \angle HAB - \angle HBA = 180^\circ - 38^\circ - 36^\circ = 106^\circ
$$

This confirms the result.

---

### Final Answer

$$
\boxed{106}
$$

[Assistant]
Explanation:
Okay, so I need to find the measure of angle AHB in triangle ABC where AD and BE are altitudes intersecting at H. They gave me angles BAC and ABC as 54 degrees and 52 degrees respectively. Let me try to visualize the triangle first.

First, let me sketch triangle ABC. Let me recall that in any triangle, the sum of the angles is 180 degrees. So, if angle BAC is 54° and angle ABC is 52°, then angle ACB can be calculated. Let me do that first.

So, angle BAC + angle ABC + angle ACB = 180°
54° + 52° + angle ACB = 180°
Adding 54 and 52 gives 106°, so angle ACB = 180° - 106° = 74°. Okay, so angle C is 74 degrees.

Now, AD and BE are altitudes. Let me recall that an altitude of a triangle is a perpendicular segment from a vertex to the opposite side (or the line containing the opposite side). So, AD is the altitude from A to BC, meaning AD is perpendicular to BC. Similarly, BE is the altitude from B to AC, so BE is perpendicular to AC.

They intersect at H, which is the orthocenter of triangle ABC. The orthocenter is the point where all three altitudes intersect. So, H is the orthocenter here.

Now, the question is asking for angle AHB. So, angle at H between points A and B. Let me think about how to find this angle.

I remember that in a triangle, the measure of the angle formed by two altitudes can be related to the angles of the triangle. Maybe there's a formula or a relationship here. Alternatively, maybe I can use some properties of cyclic quadrilaterals or triangle angle sums.

Alternatively, maybe I can find coordinates or use trigonometry. But since angles are given, perhaps there's a more straightforward geometric approach.

Let me recall that in triangle ABC, the orthocenter H creates several right angles with the sides. For example, since AD is an altitude, angle ADB is 90°, and similarly, angle BEA is 90°. Maybe I can use these right angles to find some relationships.

Alternatively, maybe I can consider quadrilateral HDCE or something like that? Wait, not sure. Let me think.

Alternatively, since H is the orthocenter, maybe I can find some angles in triangle AHB. Let me consider triangle AHB. If I can find some angles in this triangle, then I can compute angle AHB.

Let me note that in triangle ABC, angles at A, B, C are 54°, 52°, 74° respectively. Let me try to find some relationships involving H.

Since AD and BE are altitudes, then AD ⊥ BC and BE ⊥ AC. Let me consider angles involving H.

Let me think about the angles at H. For example, angle AHB. Let me recall that in some cases, the measure of angle AHB can be related to the angles of the original triangle. For example, in an acute triangle, the orthocenter lies inside the triangle, and angle AHB is equal to 180° minus angle ACB. Wait, is that true?

Wait, let me check. Let me recall that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is that a general formula?

Wait, let me think. Let me consider quadrilateral HDCE, but maybe not. Alternatively, let me consider triangle AHB. Let me try to find angles at H.

Alternatively, since AD and BE are altitudes, then angles at D and E are right angles. Let me consider triangle AHB. Let me try to find angles at A and B in triangle AHB.

Wait, angle at A in triangle AHB is the same as angle between HA and AB. Wait, but HA is part of the altitude AD. Wait, maybe not. Let me think.

Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates several similar triangles or something.

Alternatively, maybe I can use the fact that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Wait, if that's the case, then angle AHB would be 180° - 74° = 106°. But I need to verify if this is correct.

Wait, let me think again. Let me consider triangle AHB. Let me try to find angles at H. Let me think about the angles formed by the altitudes.

Since AD is an altitude, angle ADB is 90°, and BE is an altitude, angle BEA is 90°. Let me consider quadrilateral AEHD. Wait, maybe not. Let me think about triangle AHB.

Alternatively, maybe I can use the fact that in triangle ABC, the orthocenter H creates angles that are supplementary to the original angles. Wait, here's a thought: In triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Let me check this with an example.

Suppose we have an equilateral triangle, all angles 60°, then the orthocenter coincides with the centroid and the circumcenter. Then angle AHB would be... Well, in that case, all angles at H would be 120°, since each angle between two altitudes would be 120°, which is 180° - 60°, so that seems to fit. So maybe that formula is correct.

So if angle ACB is 74°, then angle AHB would be 180° - 74° = 106°. But let me verify this with another approach to be sure.

Alternatively, let me try to compute angle AHB using other methods.

Let me consider triangle AHB. Let me find angles at A and B in triangle AHB. Wait, angle at A in triangle AHB is the same as angle between HA and AB. But HA is part of the altitude from A to BC. Let me think.

Alternatively, since AD is an altitude, then AD is perpendicular to BC. Similarly, BE is perpendicular to AC. Let me consider angles involving H.

Let me consider triangle AHB. Let me denote angle HAB and angle HBA, then angle AHB = 180° - angle HAB - angle HBA.

So, if I can find angle HAB and angle HBA, then I can compute angle AHB.

Let me find angle HAB. Since AD is an altitude, angle BAD is equal to 90° - angle ABC. Wait, let me think.

In triangle ABC, angle at B is 52°, and AD is the altitude from A to BC. So, in triangle ABD, which is a right triangle (since AD is an altitude), angle at D is 90°, angle at B is 52°, so angle BAD = 90° - angle ABC = 90° - 52° = 38°. Therefore, angle BAD is 38°, which is angle between BA and AD. But angle HAB is the same as angle BAD because H lies on AD. Therefore, angle HAB is 38°.

Similarly, let me find angle HBA. Since BE is the altitude from B to AC, so in triangle BEA, which is a right triangle, angle at E is 90°, angle at A is 54°, so angle ABE = 90° - angle BAC = 90° - 54° = 36°. Therefore, angle ABE is 36°, which is the angle between BA and BE. Since H lies on BE, angle HBA is equal to angle ABE, which is 36°.

Therefore, in triangle AHB, angles at A and B are 38° and 36°, so angle AHB = 180° - 38° - 36° = 180° - 74° = 106°. So that gives angle AHB as 106°, which matches the previous thought.

Therefore, the answer is 106 degrees. Let me check again if this is correct.

Wait, let me verify the steps again. First, angle BAD is 90° - angle ABC. Is that correct?

Yes, because in triangle ABD, right-angled at D, angle at B is angle ABC = 52°, so angle BAD = 90° - 52° = 38°, correct. Similarly, angle ABE is 90° - angle BAC = 90° - 54° = 36°, correct. Then, in triangle AHB, angles at A and B are 38° and 36°, so angle AHB is 180 - 38 - 36 = 106°, which is 180 - 74°, which is 180 - angle ACB. So that seems to hold.

Alternatively, another way to think about this is that angle AHB = 180° - angle ABC - angle BAC? Wait, no, that would be 180 - 54 - 52 = 74°, which is angle ACB. But that's not the case here. Wait, but angle AHB is 106°, which is 180 - 74°, so angle AHB = 180° - angle ACB. So that seems to be a general formula.

So, in general, in any triangle, the measure of angle AHB is equal to 180° minus the measure of angle ACB. Is that a general rule?

Let me think. Let me consider another example. Suppose we have a triangle where angle C is 90°, so it's a right-angled triangle. Then the orthocenter H is at point C, because in a right-angled triangle, the altitudes are the legs themselves. So, angle AHB would be angle ACB, which is 90°, but according to the formula 180° - angle ACB, that would be 90°, which matches. Wait, but if H is at C, then angle AHB is angle ACB? Wait, if H is at C, then angle AHB is angle ACB? Wait, no. If H is at C, then points A, H, B would be A, C, B, so angle AHB is angle ACB, which is 90°, and according to the formula 180° - angle ACB would be 90°, which matches. So that works.

Another example: take an equilateral triangle, all angles 60°, then the orthocenter is at the center. Then angle AHB would be 120°, which is 180° - 60°, so that works. So seems like the formula angle AHB = 180° - angle ACB is correct.

Therefore, in this problem, since angle ACB is 74°, angle AHB is 180° - 74° = 106°, so the answer is 106 degrees.

But let me check again with the previous calculation where I found angles HAB and HBA as 38° and 36°, adding up to 74°, so angle AHB is 106°, which is consistent. Therefore, seems correct.

Alternatively, maybe I can use coordinates to verify. Let me try that.

Let me assign coordinates to triangle ABC. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Wait, but maybe it's easier to use trigonometry.

Alternatively, let me use the Law of Sines to find some relations. Wait, but maybe coordinates would be more straightforward.

Let me try coordinates. Let me assume coordinates for triangle ABC. Let me set point A at (0, 0), point B at (b, 0), and point C at coordinates (d, e). But maybe this is too vague. Alternatively, let me construct triangle ABC with given angles.

Alternatively, since angles are given, maybe I can assign lengths based on the Law of Sines.

Let me denote sides as follows: side opposite angle A is BC = a, side opposite angle B is AC = b, side opposite angle C is AB = c.

By Law of Sines:

a / sin A = b / sin B = c / sin C

Given angles:

Angle A = 54°, angle B = 52°, angle C = 74°

So, let me assign some lengths. Let me assume that the circumradius is 1, but maybe it's easier to assign specific lengths.

Alternatively, let me take AB = c, BC = a, AC = b.

But maybe coordinates. Let me place point A at (0, 0), point B at (c, 0), and point C somewhere in the plane. Let me compute coordinates based on angles.

Alternatively, since I need altitudes, maybe coordinates would complicate. Alternatively, maybe use vectors or trigonometry.

Alternatively, since I already have two methods giving me 106°, maybe I can be confident. But just to be thorough, let me try another approach.

Let me recall that in triangle ABC, the orthocenter H. Then, angles involving H can be related to the original angles. For example, in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. This is a known result. Let me try to recall why that is.

Consider the orthocenter H. Then, quadrilateral HECD, where E and D are the feet of the altitudes, is cyclic? Wait, not sure. Alternatively, since AD and BE are altitudes, intersecting at H, then angles at H can be related.

Alternatively, consider that in triangle ABC, angle HAB = 90° - angle ABC, and angle HBA = 90° - angle BAC, as we found earlier. Therefore, in triangle AHB, angles at A and B are 90° - angle ABC and 90° - angle BAC, respectively. Therefore, angle AHB = 180° - [ (90° - angle ABC) + (90° - angle BAC) ] = 180° - 180° + angle ABC + angle BAC = angle ABC + angle BAC.

Wait, hold on! Wait, that would be angle AHB = angle ABC + angle BAC. But angle ABC + angle BAC is 52° + 54° = 106°, which matches our previous result. Wait, but earlier I thought angle AHB = 180° - angle ACB, which is also 180° - 74° = 106°, so both formulas are consistent. Because angle ABC + angle BAC = 180° - angle ACB. Since angle ABC + angle BAC + angle ACB = 180°, so angle ABC + angle BAC = 180° - angle ACB. Therefore, angle AHB = angle ABC + angle BAC = 180° - angle ACB. Therefore, both formulas are equivalent.

Therefore, angle AHB is equal to the sum of angles at A and B, which is 54° + 52° = 106°, or 180° - angle C = 180° - 74° = 106°. So, both ways, same answer.

Therefore, the measure of angle AHB is 106 degrees.

But let me check once more with another method. Let me use coordinates.

Let me try to assign coordinates to triangle ABC. Let me place point A at (0, 0), point B at (1, 0), and point C somewhere in the plane. Let me compute coordinates of C based on angles.

But maybe it's easier to use Law of Sines. Let me assume AB = c, BC = a, AC = b.

From Law of Sines:

a / sin A = b / sin B = c / sin C

Let me assign AB = c = 1 (for simplicity). Then:

a / sin 54° = 1 / sin 74°, so a = sin 54° / sin 74°

Similarly, b / sin 52° = 1 / sin 74°, so b = sin 52° / sin 74°

Let me compute these values numerically.

First, compute sin 54°, sin 52°, sin 74°.

Using calculator:

sin 54° ≈ 0.8090

sin 52° ≈ 0.7880

sin 74° ≈ 0.9613

Therefore:

a ≈ 0.8090 / 0.9613 ≈ 0.8416

b ≈ 0.7880 / 0.9613 ≈ 0.8197

So, sides:

AB = 1

BC ≈ 0.8416

AC ≈ 0.8197

Now, let me assign coordinates. Let me place point A at (0, 0), point B at (1, 0). Now, need to find coordinates of point C.

Let me denote coordinates of C as (x, y). Then, distance from A to C is b ≈ 0.8197, so:

x² + y² = b² ≈ (0.8197)² ≈ 0.6719

Distance from B to C is a ≈ 0.8416, so:

(x - 1)² + y² = a² ≈ (0.8416)² ≈ 0.7083

Subtracting first equation from the second:

(x - 1)² + y² - x² - y² = 0.7083 - 0.6719

Expanding (x - 1)² - x²:

x² - 2x + 1 - x² = -2x + 1 = 0.7083 - 0.6719 ≈ 0.0364

Therefore:

-2x + 1 ≈ 0.0364 => -2x ≈ -0.9636 => x ≈ 0.4818

Then, from x² + y² ≈ 0.6719:

(0.4818)² + y² ≈ 0.6719 => 0.2321 + y² ≈ 0.6719 => y² ≈ 0.4398 => y ≈ √0.4398 ≈ 0.6632

Therefore, coordinates of C are approximately (0.4818, 0.6632)

Now, let me find equations of altitudes AD and BE, then find their intersection H, then compute angle AHB.

First, find altitude AD: from A(0,0) perpendicular to BC.

First, find slope of BC. Coordinates of B(1, 0) and C(0.4818, 0.6632). The slope of BC is (0.6632 - 0)/(0.4818 - 1) = 0.6632 / (-0.5182) ≈ -1.279

Therefore, the slope of BC is approximately -1.279, so the slope of altitude AD, which is perpendicular to BC, is the negative reciprocal: 1 / 1.279 ≈ 0.781

Therefore, equation of altitude AD: passes through A(0,0), slope ≈ 0.781, so equation is y = 0.781x

Now, find equation of altitude BE: from B(1, 0) perpendicular to AC.

First, find slope of AC. Coordinates of A(0,0) and C(0.4818, 0.6632). Slope of AC is (0.6632 - 0)/(0.4818 - 0) ≈ 0.6632 / 0.4818 ≈ 1.376

Therefore, slope of AC ≈ 1.376, so slope of altitude BE, which is perpendicular, is -1 / 1.376 ≈ -0.727

Equation of altitude BE: passes through B(1, 0), slope ≈ -0.727. So equation is y - 0 = -0.727(x - 1), so y = -0.727x + 0.727

Now, find intersection point H of AD and BE.

AD: y = 0.781x

BE: y = -0.727x + 0.727

Set equal:

0.781x = -0.727x + 0.727

0.781x + 0.727x = 0.727

(0.781 + 0.727)x = 0.727

1.508x ≈ 0.727 => x ≈ 0.727 / 1.508 ≈ 0.482

Then y ≈ 0.781 * 0.482 ≈ 0.376

Therefore, coordinates of H are approximately (0.482, 0.376)

Now, need to compute angle AHB. Points A(0,0), H(0.482, 0.376), B(1, 0)

Compute vectors HA and HB.

Vector HA = A - H = (0 - 0.482, 0 - 0.376) = (-0.482, -0.376)

Vector HB = B - H = (1 - 0.482, 0 - 0.376) = (0.518, -0.376)

Now, angle between vectors HA and HB is angle AHB.

Use the dot product formula:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB = (-0.482)(0.518) + (-0.376)(-0.376) ≈ (-0.249) + (0.141) ≈ -0.108

Compute |HA| = sqrt((-0.482)^2 + (-0.376)^2) ≈ sqrt(0.232 + 0.141) ≈ sqrt(0.373) ≈ 0.611

Compute |HB| = sqrt((0.518)^2 + (-0.376)^2) ≈ sqrt(0.268 + 0.141) ≈ sqrt(0.409) ≈ 0.640

Therefore, cos(theta) ≈ -0.108 / (0.611 * 0.640) ≈ -0.108 / 0.391 ≈ -0.276

Therefore, theta ≈ arccos(-0.276) ≈ 106 degrees (since cos(106°) ≈ cos(90° + 16°) = -sin(16°) ≈ -0.2756). Yes! That's very close. So arccos(-0.276) is approximately 106 degrees. Therefore, angle AHB is approximately 106°, which matches our previous results.

Therefore, this coordinate method also gives 106°, so I can be confident that the answer is 106 degrees.

Therefore, the measure of angle AHB is 106 degrees.

**Final Answer**
\boxed{106}

To determine the measure of angle $ \angle AHB $ in triangle $ \triangle ABC $, where altitudes $ \overline{AD} $ and $ \overline{BE} $ intersect at the orthocenter $ H $, we are given the following:

- $ \angle BAC = 54^\circ $
- $ \angle ABC = 52^\circ $

---

### Step 1: Determine the Third Angle of Triangle $ ABC $

In any triangle, the sum of the interior angles is $ 180^\circ $. Therefore, we can compute the third angle:

$$
\angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 54^\circ - 52^\circ = 74^\circ
$$

---

### Step 2: Use a Known Geometric Property

A key geometric property of the orthocenter $ H $ in triangle $ ABC $ is that the angle formed at the orthocenter between two vertices (e.g., $ \angle AHB $) is related to the third angle of the triangle. Specifically:

$$
\angle AHB = 180^\circ - \angle ACB
$$

This is because the orthocenter lies inside the triangle (since all angles are acute), and the angle $ \angle AHB $ is formed by two altitudes intersecting at $ H $, which creates a supplementary angle with $ \angle ACB $.

---

### Step 3: Apply the Formula

Using the known value of $ \angle ACB = 74^\circ $, we compute:

$$
\angle AHB = 180^\circ - 74^\circ = 106^\circ
$$

---

### Step 4: Verification via Triangle Angle Sum

Alternatively, we can verify this by considering triangle $ \triangle AHB $. The angles at $ A $ and $ B $ in this triangle are formed by the altitudes:

- $ \angle HAB = 90^\circ - \angle ABC = 90^\circ - 52^\circ = 38^\circ $
- $ \angle HBA = 90^\circ - \angle BAC = 90^\circ - 54^\circ = 36^\circ $

Then, in triangle $ \triangle AHB $, the angle at $ H $ is:

$$
\angle AHB = 180^\circ - \angle HAB - \angle HBA = 180^\circ - 38^\circ - 36^\circ = 106^\circ
$$

This confirms the result.

---

### Final Answer

$$
\boxed{106}
$$

Solution:
First, we build a diagram:

[asy]

size(150); defaultpen(linewidth(0.8));

pair B = (0,0), C = (3,0), A = (1.8,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);

draw(A--B--C--cycle);

draw(A--P^^B--Q);

label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,E); label("$H$",H,NW);

draw(rightanglemark(C,P,H,3.5));

draw(rightanglemark(H,Q,C,3.5));

[/asy]

We have $\angle AHB = \angle DHE$, and from quadrilateral $CDHE$, we have  \begin{align*}
\angle DHE &= 360^\circ - \angle HEC - \angle ECD - \angle CDH \\
&= 360^\circ - 90^\circ - \angle ACB - 90^\circ\\
&= 180^\circ - \angle ACB.
\end{align*}From triangle $ABC$, we have $180^\circ - \angle ACB = \angle BAC + \angle ABC = 54^\circ + 52^\circ = \boxed{106^\circ}$.