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 the problem gives angles, perhaps there's a way to find angle AHB using angle chasing.

Let me try angle chasing. Let me recall that in triangle ABC, with altitudes AD and BE intersecting at H. Let me consider quadrilateral HDCE or something? Wait, maybe not. Let me think.

Since AD is an altitude, angle ADC is 90°, and BE is an altitude, so angle BEA is 90°. Maybe I can find some right triangles here.

Alternatively, since H is the orthocenter, maybe I can relate angles at H with the original angles of the triangle.

I remember that in a triangle, the measure of the angle between two altitudes can be calculated as 180° minus the measure of the original angle. Wait, let me think again.

Alternatively, maybe angle AHB is equal to 180° minus angle ACB? Wait, let me check with some known cases. For example, in an acute triangle, the orthocenter lies inside the triangle. Let me consider a specific case. Suppose triangle ABC is such that angle C is 90°, then the orthocenter is at point C, so angle AHB would be angle ACB? Wait, no. If angle C is 90°, then altitudes from A and B would be AC and BC respectively, so their intersection is at C. So angle AHB would be angle ACB, which is 90°, which is equal to 180° - angle ACB? Wait, 180° - 90° is 90°, so that's equal. Wait, maybe angle AHB = 180° - angle ACB?

Wait, let me check another case. Suppose angle C is 60°, then angle AHB would be 180° - 60° = 120°? Let me see. If I have triangle ABC with angle C = 60°, then the orthocenter H... Hmm, maybe I need to calculate it.

Alternatively, maybe there's a general formula. Let me think. In triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is that true?

Wait, let me try to derive it. Let me consider triangle AHB. If I can find the other angles in triangle AHB, then I can find angle AHB.

Alternatively, since AD and BE are altitudes, then HD is perpendicular to BC and HE is perpendicular to AC. Maybe I can find some relationships.

Alternatively, let me consider quadrilateral AEHF or something. Wait, maybe not. Let me think again.

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 from A. Wait, maybe not. Let me think.

Alternatively, since AD is an altitude, then AD is perpendicular to BC. Similarly, BE is perpendicular to AC. So, maybe I can find some angles in triangle AHB by considering the right angles.

Let me try to find angle HAB and angle HBA, then angle AHB would be 180° minus those two.

So, angle HAB is the angle between HA and AB. Since HA is part of the altitude from A, which is AD. Wait, HA is along AD? Wait, no. AD is the altitude from A to BC, so AD is perpendicular to BC. But H is the intersection of AD and BE. So HA is part of AD. Therefore, angle HAB is the angle between HA (which is along AD) and AB.

Similarly, angle HBA is the angle between HB (which is along BE) and BA.

So, maybe I can find angle HAB and angle HBA by using the fact that AD and BE are altitudes.

Let me try to find angle HAB. Since AD is an altitude, angle BAD is equal to 90° minus angle ABC. Wait, let me think. In triangle ABD, which is a right triangle at D, angle BAD + angle ABD = 90°. Wait, angle ABD is angle ABC, which is 52°, so angle BAD = 90° - 52° = 38°. Therefore, angle BAD is 38°, which is angle between BA and AD. But angle HAB is the same as angle BAD, since HA is along AD. Therefore, angle HAB is 38°.

Similarly, let me find angle HBA. Since BE is an altitude, angle ABE is equal to 90° minus angle BAC. Wait, let me check. In triangle ABE, which is a right triangle at E, angle ABE + angle BAE = 90°. Angle BAE is angle BAC, which is 54°, so angle ABE = 90° - 54° = 36°. Therefore, angle ABE is 36°, which is the angle between BA and BE. Since HB is along 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 is 180° - 38° - 36° = 180° - 74° = 106°. Wait, but angle ACB was 74°, so angle AHB is 180° - angle ACB? Which would be 180° - 74° = 106°, which matches. So that seems to be the case.

Therefore, angle AHB is 180° minus angle ACB. Therefore, since angle ACB is 74°, angle AHB is 106°. Therefore, the answer is 106 degrees. Let me check if this is correct.

Wait, let me verify again. So angle BAD is 90° - angle ABC. Since angle ABC is 52°, angle BAD is 90 - 52 = 38°, which is angle HAB. Similarly, angle ABE is 90° - angle BAC = 90 - 54 = 36°, which is angle HBA. Then angle AHB is 180 - 38 - 36 = 106°, which is 180 - 74, which is 106. So that seems correct.

Alternatively, another way to think about it is that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. So, if we can remember that formula, then we can just compute 180 - 74 = 106°. But I need to make sure that this formula is correct.

Alternatively, let me think of another approach. Since H is the orthocenter, then angles involving H can be related to the original angles. For example, in triangle ABC, the orthocenter creates several right angles. Also, quadrilaterals formed by the orthocenter and the vertices can be cyclic? Wait, not sure.

Alternatively, since AD and BE are altitudes, then angles AHB and ACB are related. Let me think about cyclic quadrilaterals. If I consider quadrilateral HECD or something. Wait, maybe not. Alternatively, since angles at D and E are right angles, maybe there's some cyclic quadrilateral.

Alternatively, since angle HAB = 90° - angle ABC and angle HBA = 90° - angle BAC, then angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = 180° - 90° + angle ABC - 90° + angle BAC = angle ABC + angle BAC. Wait, that would be angle AHB = angle BAC + angle ABC. Wait, but angle BAC + angle ABC is 54 + 52 = 106°, which is the same as before. Wait, so angle AHB = angle BAC + angle ABC? But angle BAC + angle ABC is 106°, which is equal to 180° - angle ACB. Since angle ACB is 74°, so 180 - 74 = 106. So both ways, same result.

Therefore, angle AHB = angle BAC + angle ABC. Wait, but why is that?

Wait, let me check again. If angle AHB = 180° - angle HAB - angle HBA. And angle HAB = 90° - angle ABC, angle HBA = 90° - angle BAC. Therefore, angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = 180° - 90° + angle ABC - 90° + angle BAC = angle ABC + angle BAC. So yes, angle AHB = angle BAC + angle ABC. Which is 54 + 52 = 106°, which is the answer.

Alternatively, since angle AHB = 180° - angle ACB, which is 180 - 74 = 106°, same result. Therefore, both methods give the same answer. Therefore, seems like the answer is 106 degrees.

But let me check with another approach. Let me try to use coordinates. Maybe assign coordinates to the triangle and calculate the coordinates of H, then compute the angle.

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

Alternatively, since I know all angles, maybe I can assign lengths. Let me assume some lengths for sides to compute coordinates.

But maybe that's too time-consuming. Alternatively, since I already have two methods giving me 106°, maybe I can be confident. But let me think again.

Wait, another way: In triangle ABC, the orthocenter H. Then, angles involving H. For example, in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is this a general formula?

Yes, actually, I think this is a known property. In any triangle, the measure of the angle formed by two altitudes (i.e., angle AHB) is equal to 180° minus the measure of the original angle at C. So, angle AHB = 180° - angle ACB. Therefore, since angle ACB is 74°, angle AHB is 106°, which is the answer.

Alternatively, if I recall, in triangle ABC, the orthocenter H, then:

angle AHB = 180° - angle ACB

angle BHC = 180° - angle BAC

angle AHC = 180° - angle ABC

So, this seems to be a standard result. Therefore, if I remember this, then angle AHB is 180° - angle ACB = 180 - 74 = 106°, so answer is 106°.

But to make sure, let me think again why angle AHB = 180° - angle ACB.

Let me consider quadrilateral HEBD, where D and E are the feet of the altitudes. Wait, maybe not. Alternatively, since AD and BE are altitudes, then HD is perpendicular to BC and HE is perpendicular to AC. Maybe considering cyclic quadrilaterals.

Alternatively, since angle HAB = 90° - angle ABC and angle HBA = 90° - angle BAC, then angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = angle ABC + angle BAC. Which is equal to 180° - angle ACB, since angle ABC + angle BAC + angle ACB = 180°, so angle ABC + angle BAC = 180° - angle ACB. Therefore, angle AHB = 180° - angle ACB. Therefore, this is a general formula.

Therefore, regardless of the specific triangle, angle AHB is equal to 180° minus angle ACB. Therefore, in this case, since angle ACB is 74°, angle AHB is 106°, so the answer is 106 degrees.

Therefore, I think that is the correct answer. Let me check once again with actual calculation.

Given angles at A and B are 54° and 52°, so angle at C is 74°, as calculated. Then angle AHB = 180 - 74 = 106°, so answer is 106°. Therefore, I think that is correct.

Alternatively, if I want to check with coordinates, let me try.

Let me place triangle ABC with coordinates. Let me assume coordinates for simplicity. Let me set point A at (0, 0), point B at (b, 0), and point C at coordinates (c, d). Then, I can compute equations for altitudes AD and BE, find their intersection H, then compute angle AHB.

But this might take some time, but let me try.

First, let me compute coordinates. Let me assume some lengths. Let me suppose AB is of length 1 unit for simplicity. Wait, but maybe it's better to use actual angles. Alternatively, use the Law of Sines to find ratios of sides.

In triangle ABC, angles are:

Angle at A: 54°, angle at B: 52°, angle at C: 74°.

By Law of Sines:

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

Where a is BC, b is AC, c is AB.

Let me denote:

Let me denote sides as follows:

Side opposite angle A (which is BC) is a.

Side opposite angle B (which is AC) is b.

Side opposite angle C (which is AB) is c.

Therefore:

a / sin 54° = b / sin 52° = c / sin 74°

Let me assign a specific length. Let me set c = AB = 1. Then:

c = 1 = AB.

Therefore, a = BC = (sin 54° / sin 74°) * c ≈ (0.8090 / 0.9613) * 1 ≈ 0.8416

Similarly, b = AC = (sin 52° / sin 74°) * c ≈ (0.7880 / 0.9613) * 1 ≈ 0.820

So, sides:

AB = 1

BC ≈ 0.8416

AC ≈ 0.820

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.

Since AC ≈ 0.820 and angle at A is 54°, so coordinates of C can be found using polar coordinates from A. Wait, but need to consider coordinates.

Alternatively, since AB is from (0,0) to (1,0). Let me find coordinates of C such that AC ≈ 0.820 and angle at A is 54°. Wait, angle at A is between AB and AC. So, if AB is along x-axis from (0,0) to (1,0), then AC makes an angle of 54° with AB. Therefore, coordinates of C can be (AC * cos(54°), AC * sin(54°)) ≈ (0.820 * cos(54°), 0.820 * sin(54°)).

Calculating:

cos(54°) ≈ 0.5878

sin(54°) ≈ 0.8090

Therefore, coordinates of C ≈ (0.820 * 0.5878, 0.820 * 0.8090) ≈ (0.481, 0.663)

But we also need to ensure that BC ≈ 0.8416. Let me check the distance between B(1, 0) and C(0.481, 0.663):

Distance BC = sqrt[(1 - 0.481)^2 + (0 - 0.663)^2] ≈ sqrt[(0.519)^2 + (-0.663)^2] ≈ sqrt[0.269 + 0.439] ≈ sqrt[0.708] ≈ 0.841, which matches our earlier calculation. Good.

So coordinates:

A(0, 0)

B(1, 0)

C(≈0.481, ≈0.663)

Now, need to find equations of altitudes AD and BE.

First, altitude from A to BC: AD.

First, find equation of BC to find the foot D.

Coordinates of B(1, 0) and C(0.481, 0.663). Let me compute the slope of BC.

Slope of BC: (0.663 - 0)/(0.481 - 1) = 0.663 / (-0.519) ≈ -1.278

Therefore, the line BC has slope m1 ≈ -1.278. Therefore, the altitude from A to BC is perpendicular to BC, so its slope is the negative reciprocal: m_AD = 1 / 1.278 ≈ 0.782.

Therefore, equation of altitude AD: passes through A(0, 0), slope ≈ 0.782. So equation is y = 0.782x.

Now, equation of BC: Let me write it in exact terms. Let me use exact coordinates.

But maybe better to compute exact equations. Alternatively, let me use exact expressions.

Alternatively, since I have approximate coordinates, let me proceed with approximate values.

Equation of BC: Let me write it in point-slope form. Let me take point B(1, 0). The slope is m1 ≈ -1.278. Therefore, equation is y - 0 = m1(x - 1), so y ≈ -1.278(x - 1).

Equation of altitude AD: y ≈ 0.782x.

Find intersection point D between BC and AD.

Set 0.782x ≈ -1.278(x - 1)

0.782x ≈ -1.278x + 1.278

0.782x + 1.278x ≈ 1.278

2.06x ≈ 1.278

x ≈ 1.278 / 2.06 ≈ 0.619

Then y ≈ 0.782 * 0.619 ≈ 0.483

Therefore, coordinates of D ≈ (0.619, 0.483)

Now, altitude BE: from B to AC.

First, find equation of AC. Coordinates of A(0, 0) and C(0.481, 0.663). Slope of AC is (0.663 - 0)/(0.481 - 0) ≈ 0.663 / 0.481 ≈ 1.378.

Therefore, slope of AC ≈ 1.378. Therefore, the altitude from B to AC is perpendicular, so slope is -1 / 1.378 ≈ -0.726.

Equation of altitude BE: passes through B(1, 0), slope ≈ -0.726. So equation is y - 0 = -0.726(x - 1), so y ≈ -0.726x + 0.726.

Now, find intersection point E between BE and AC.

Equation of AC: passes through A(0, 0) with slope ≈1.378, so equation is y ≈ 1.378x.

Set equations equal:

1.378x ≈ -0.726x + 0.726

1.378x + 0.726x ≈ 0.726

2.104x ≈ 0.726

x ≈ 0.726 / 2.104 ≈ 0.345

Then y ≈ 1.378 * 0.345 ≈ 0.475

Therefore, coordinates of E ≈ (0.345, 0.475)

Now, we have equations of AD: y ≈ 0.782x and BE: y ≈ -0.726x + 0.726. Their intersection is point H.

Solve for x and y:

0.782x = -0.726x + 0.726

0.782x + 0.726x = 0.726

1.508x ≈ 0.726

x ≈ 0.726 / 1.508 ≈ 0.4816

Then y ≈ 0.782 * 0.4816 ≈ 0.376

Therefore, coordinates of H ≈ (0.4816, 0.376)

Now, we need to find angle AHB. Points A(0, 0), H(0.4816, 0.376), B(1, 0). Let me compute vectors HA and HB, then compute the angle between them.

Vector HA = A - H = (0 - 0.4816, 0 - 0.376) = (-0.4816, -0.376)

Vector HB = B - H = (1 - 0.4816, 0 - 0.376) = (0.5184, -0.376)

Now, angle between vectors HA and HB can be found using the dot product formula:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB = (-0.4816)(0.5184) + (-0.376)(-0.376)

First term: -0.4816 * 0.5184 ≈ -0.250

Second term: 0.376^2 ≈ 0.141

Total ≈ -0.250 + 0.141 ≈ -0.109

Now compute |HA|: sqrt((-0.4816)^2 + (-0.376)^2) ≈ sqrt(0.2319 + 0.1414) ≈ sqrt(0.3733) ≈ 0.611

Compute |HB|: sqrt((0.5184)^2 + (-0.376)^2) ≈ sqrt(0.2688 + 0.1414) ≈ sqrt(0.4102) ≈ 0.6405

Therefore, cos(theta) ≈ -0.109 / (0.611 * 0.6405) ≈ -0.109 / 0.391 ≈ -0.279

Therefore, theta ≈ arccos(-0.279) ≈ 106.3 degrees

Which is approximately 106 degrees, which matches our previous result. Therefore, this numerical calculation confirms that angle AHB is approximately 106 degrees. Therefore, the answer is 106°, so boxed answer is \boxed{106}.

**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 begin by analyzing the given information and applying geometric properties.

---

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

We are given:
- $ \angle BAC = 54^\circ $
- $ \angle ABC = 52^\circ $

Using the triangle angle sum property:
$$
\angle BAC + \angle ABC + \angle ACB = 180^\circ
$$
$$
54^\circ + 52^\circ + \angle ACB = 180^\circ
$$
$$
\angle ACB = 180^\circ - 106^\circ = 74^\circ
$$

---

### Step 2: Use the Property of the Orthocenter

In any triangle, the **orthocenter** $ H $ is the point where the three altitudes intersect. A key geometric property is that the angle formed at the orthocenter by two altitudes (e.g., $ \angle AHB $) is related to the third angle of the triangle.

Specifically, the following identity holds:
$$
\angle AHB = 180^\circ - \angle ACB
$$

This is because the altitudes form right angles with the opposite sides, and the triangle formed by the orthocenter and two vertices has angles that are supplementary to the original triangle's angles.

---

### Step 3: Apply the Formula

We already found:
$$
\angle ACB = 74^\circ
$$

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

---

### 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 the problem gives angles, perhaps there's a way to find angle AHB using angle chasing.

Let me try angle chasing. Let me recall that in triangle ABC, with altitudes AD and BE intersecting at H. Let me consider quadrilateral HDCE or something? Wait, maybe not. Let me think.

Since AD is an altitude, angle ADC is 90°, and BE is an altitude, so angle BEA is 90°. Maybe I can find some right triangles here.

Alternatively, since H is the orthocenter, maybe I can relate angles at H with the original angles of the triangle.

I remember that in a triangle, the measure of the angle between two altitudes can be calculated as 180° minus the measure of the original angle. Wait, let me think again.

Alternatively, maybe angle AHB is equal to 180° minus angle ACB? Wait, let me check with some known cases. For example, in an acute triangle, the orthocenter lies inside the triangle. Let me consider a specific case. Suppose triangle ABC is such that angle C is 90°, then the orthocenter is at point C, so angle AHB would be angle ACB? Wait, no. If angle C is 90°, then altitudes from A and B would be AC and BC respectively, so their intersection is at C. So angle AHB would be angle ACB, which is 90°, which is equal to 180° - angle ACB? Wait, 180° - 90° is 90°, so that's equal. Wait, maybe angle AHB = 180° - angle ACB?

Wait, let me check another case. Suppose angle C is 60°, then angle AHB would be 180° - 60° = 120°? Let me see. If I have triangle ABC with angle C = 60°, then the orthocenter H... Hmm, maybe I need to calculate it.

Alternatively, maybe there's a general formula. Let me think. In triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is that true?

Wait, let me try to derive it. Let me consider triangle AHB. If I can find the other angles in triangle AHB, then I can find angle AHB.

Alternatively, since AD and BE are altitudes, then HD is perpendicular to BC and HE is perpendicular to AC. Maybe I can find some relationships.

Alternatively, let me consider quadrilateral AEHF or something. Wait, maybe not. Let me think again.

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 from A. Wait, maybe not. Let me think.

Alternatively, since AD is an altitude, then AD is perpendicular to BC. Similarly, BE is perpendicular to AC. So, maybe I can find some angles in triangle AHB by considering the right angles.

Let me try to find angle HAB and angle HBA, then angle AHB would be 180° minus those two.

So, angle HAB is the angle between HA and AB. Since HA is part of the altitude from A, which is AD. Wait, HA is along AD? Wait, no. AD is the altitude from A to BC, so AD is perpendicular to BC. But H is the intersection of AD and BE. So HA is part of AD. Therefore, angle HAB is the angle between HA (which is along AD) and AB.

Similarly, angle HBA is the angle between HB (which is along BE) and BA.

So, maybe I can find angle HAB and angle HBA by using the fact that AD and BE are altitudes.

Let me try to find angle HAB. Since AD is an altitude, angle BAD is equal to 90° minus angle ABC. Wait, let me think. In triangle ABD, which is a right triangle at D, angle BAD + angle ABD = 90°. Wait, angle ABD is angle ABC, which is 52°, so angle BAD = 90° - 52° = 38°. Therefore, angle BAD is 38°, which is angle between BA and AD. But angle HAB is the same as angle BAD, since HA is along AD. Therefore, angle HAB is 38°.

Similarly, let me find angle HBA. Since BE is an altitude, angle ABE is equal to 90° minus angle BAC. Wait, let me check. In triangle ABE, which is a right triangle at E, angle ABE + angle BAE = 90°. Angle BAE is angle BAC, which is 54°, so angle ABE = 90° - 54° = 36°. Therefore, angle ABE is 36°, which is the angle between BA and BE. Since HB is along 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 is 180° - 38° - 36° = 180° - 74° = 106°. Wait, but angle ACB was 74°, so angle AHB is 180° - angle ACB? Which would be 180° - 74° = 106°, which matches. So that seems to be the case.

Therefore, angle AHB is 180° minus angle ACB. Therefore, since angle ACB is 74°, angle AHB is 106°. Therefore, the answer is 106 degrees. Let me check if this is correct.

Wait, let me verify again. So angle BAD is 90° - angle ABC. Since angle ABC is 52°, angle BAD is 90 - 52 = 38°, which is angle HAB. Similarly, angle ABE is 90° - angle BAC = 90 - 54 = 36°, which is angle HBA. Then angle AHB is 180 - 38 - 36 = 106°, which is 180 - 74, which is 106. So that seems correct.

Alternatively, another way to think about it is that in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. So, if we can remember that formula, then we can just compute 180 - 74 = 106°. But I need to make sure that this formula is correct.

Alternatively, let me think of another approach. Since H is the orthocenter, then angles involving H can be related to the original angles. For example, in triangle ABC, the orthocenter creates several right angles. Also, quadrilaterals formed by the orthocenter and the vertices can be cyclic? Wait, not sure.

Alternatively, since AD and BE are altitudes, then angles AHB and ACB are related. Let me think about cyclic quadrilaterals. If I consider quadrilateral HECD or something. Wait, maybe not. Alternatively, since angles at D and E are right angles, maybe there's some cyclic quadrilateral.

Alternatively, since angle HAB = 90° - angle ABC and angle HBA = 90° - angle BAC, then angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = 180° - 90° + angle ABC - 90° + angle BAC = angle ABC + angle BAC. Wait, that would be angle AHB = angle BAC + angle ABC. Wait, but angle BAC + angle ABC is 54 + 52 = 106°, which is the same as before. Wait, so angle AHB = angle BAC + angle ABC? But angle BAC + angle ABC is 106°, which is equal to 180° - angle ACB. Since angle ACB is 74°, so 180 - 74 = 106. So both ways, same result.

Therefore, angle AHB = angle BAC + angle ABC. Wait, but why is that?

Wait, let me check again. If angle AHB = 180° - angle HAB - angle HBA. And angle HAB = 90° - angle ABC, angle HBA = 90° - angle BAC. Therefore, angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = 180° - 90° + angle ABC - 90° + angle BAC = angle ABC + angle BAC. So yes, angle AHB = angle BAC + angle ABC. Which is 54 + 52 = 106°, which is the answer.

Alternatively, since angle AHB = 180° - angle ACB, which is 180 - 74 = 106°, same result. Therefore, both methods give the same answer. Therefore, seems like the answer is 106 degrees.

But let me check with another approach. Let me try to use coordinates. Maybe assign coordinates to the triangle and calculate the coordinates of H, then compute the angle.

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

Alternatively, since I know all angles, maybe I can assign lengths. Let me assume some lengths for sides to compute coordinates.

But maybe that's too time-consuming. Alternatively, since I already have two methods giving me 106°, maybe I can be confident. But let me think again.

Wait, another way: In triangle ABC, the orthocenter H. Then, angles involving H. For example, in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Is this a general formula?

Yes, actually, I think this is a known property. In any triangle, the measure of the angle formed by two altitudes (i.e., angle AHB) is equal to 180° minus the measure of the original angle at C. So, angle AHB = 180° - angle ACB. Therefore, since angle ACB is 74°, angle AHB is 106°, which is the answer.

Alternatively, if I recall, in triangle ABC, the orthocenter H, then:

angle AHB = 180° - angle ACB

angle BHC = 180° - angle BAC

angle AHC = 180° - angle ABC

So, this seems to be a standard result. Therefore, if I remember this, then angle AHB is 180° - angle ACB = 180 - 74 = 106°, so answer is 106°.

But to make sure, let me think again why angle AHB = 180° - angle ACB.

Let me consider quadrilateral HEBD, where D and E are the feet of the altitudes. Wait, maybe not. Alternatively, since AD and BE are altitudes, then HD is perpendicular to BC and HE is perpendicular to AC. Maybe considering cyclic quadrilaterals.

Alternatively, since angle HAB = 90° - angle ABC and angle HBA = 90° - angle BAC, then angle AHB = 180° - (90° - angle ABC) - (90° - angle BAC) = angle ABC + angle BAC. Which is equal to 180° - angle ACB, since angle ABC + angle BAC + angle ACB = 180°, so angle ABC + angle BAC = 180° - angle ACB. Therefore, angle AHB = 180° - angle ACB. Therefore, this is a general formula.

Therefore, regardless of the specific triangle, angle AHB is equal to 180° minus angle ACB. Therefore, in this case, since angle ACB is 74°, angle AHB is 106°, so the answer is 106 degrees.

Therefore, I think that is the correct answer. Let me check once again with actual calculation.

Given angles at A and B are 54° and 52°, so angle at C is 74°, as calculated. Then angle AHB = 180 - 74 = 106°, so answer is 106°. Therefore, I think that is correct.

Alternatively, if I want to check with coordinates, let me try.

Let me place triangle ABC with coordinates. Let me assume coordinates for simplicity. Let me set point A at (0, 0), point B at (b, 0), and point C at coordinates (c, d). Then, I can compute equations for altitudes AD and BE, find their intersection H, then compute angle AHB.

But this might take some time, but let me try.

First, let me compute coordinates. Let me assume some lengths. Let me suppose AB is of length 1 unit for simplicity. Wait, but maybe it's better to use actual angles. Alternatively, use the Law of Sines to find ratios of sides.

In triangle ABC, angles are:

Angle at A: 54°, angle at B: 52°, angle at C: 74°.

By Law of Sines:

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

Where a is BC, b is AC, c is AB.

Let me denote:

Let me denote sides as follows:

Side opposite angle A (which is BC) is a.

Side opposite angle B (which is AC) is b.

Side opposite angle C (which is AB) is c.

Therefore:

a / sin 54° = b / sin 52° = c / sin 74°

Let me assign a specific length. Let me set c = AB = 1. Then:

c = 1 = AB.

Therefore, a = BC = (sin 54° / sin 74°) * c ≈ (0.8090 / 0.9613) * 1 ≈ 0.8416

Similarly, b = AC = (sin 52° / sin 74°) * c ≈ (0.7880 / 0.9613) * 1 ≈ 0.820

So, sides:

AB = 1

BC ≈ 0.8416

AC ≈ 0.820

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.

Since AC ≈ 0.820 and angle at A is 54°, so coordinates of C can be found using polar coordinates from A. Wait, but need to consider coordinates.

Alternatively, since AB is from (0,0) to (1,0). Let me find coordinates of C such that AC ≈ 0.820 and angle at A is 54°. Wait, angle at A is between AB and AC. So, if AB is along x-axis from (0,0) to (1,0), then AC makes an angle of 54° with AB. Therefore, coordinates of C can be (AC * cos(54°), AC * sin(54°)) ≈ (0.820 * cos(54°), 0.820 * sin(54°)).

Calculating:

cos(54°) ≈ 0.5878

sin(54°) ≈ 0.8090

Therefore, coordinates of C ≈ (0.820 * 0.5878, 0.820 * 0.8090) ≈ (0.481, 0.663)

But we also need to ensure that BC ≈ 0.8416. Let me check the distance between B(1, 0) and C(0.481, 0.663):

Distance BC = sqrt[(1 - 0.481)^2 + (0 - 0.663)^2] ≈ sqrt[(0.519)^2 + (-0.663)^2] ≈ sqrt[0.269 + 0.439] ≈ sqrt[0.708] ≈ 0.841, which matches our earlier calculation. Good.

So coordinates:

A(0, 0)

B(1, 0)

C(≈0.481, ≈0.663)

Now, need to find equations of altitudes AD and BE.

First, altitude from A to BC: AD.

First, find equation of BC to find the foot D.

Coordinates of B(1, 0) and C(0.481, 0.663). Let me compute the slope of BC.

Slope of BC: (0.663 - 0)/(0.481 - 1) = 0.663 / (-0.519) ≈ -1.278

Therefore, the line BC has slope m1 ≈ -1.278. Therefore, the altitude from A to BC is perpendicular to BC, so its slope is the negative reciprocal: m_AD = 1 / 1.278 ≈ 0.782.

Therefore, equation of altitude AD: passes through A(0, 0), slope ≈ 0.782. So equation is y = 0.782x.

Now, equation of BC: Let me write it in exact terms. Let me use exact coordinates.

But maybe better to compute exact equations. Alternatively, let me use exact expressions.

Alternatively, since I have approximate coordinates, let me proceed with approximate values.

Equation of BC: Let me write it in point-slope form. Let me take point B(1, 0). The slope is m1 ≈ -1.278. Therefore, equation is y - 0 = m1(x - 1), so y ≈ -1.278(x - 1).

Equation of altitude AD: y ≈ 0.782x.

Find intersection point D between BC and AD.

Set 0.782x ≈ -1.278(x - 1)

0.782x ≈ -1.278x + 1.278

0.782x + 1.278x ≈ 1.278

2.06x ≈ 1.278

x ≈ 1.278 / 2.06 ≈ 0.619

Then y ≈ 0.782 * 0.619 ≈ 0.483

Therefore, coordinates of D ≈ (0.619, 0.483)

Now, altitude BE: from B to AC.

First, find equation of AC. Coordinates of A(0, 0) and C(0.481, 0.663). Slope of AC is (0.663 - 0)/(0.481 - 0) ≈ 0.663 / 0.481 ≈ 1.378.

Therefore, slope of AC ≈ 1.378. Therefore, the altitude from B to AC is perpendicular, so slope is -1 / 1.378 ≈ -0.726.

Equation of altitude BE: passes through B(1, 0), slope ≈ -0.726. So equation is y - 0 = -0.726(x - 1), so y ≈ -0.726x + 0.726.

Now, find intersection point E between BE and AC.

Equation of AC: passes through A(0, 0) with slope ≈1.378, so equation is y ≈ 1.378x.

Set equations equal:

1.378x ≈ -0.726x + 0.726

1.378x + 0.726x ≈ 0.726

2.104x ≈ 0.726

x ≈ 0.726 / 2.104 ≈ 0.345

Then y ≈ 1.378 * 0.345 ≈ 0.475

Therefore, coordinates of E ≈ (0.345, 0.475)

Now, we have equations of AD: y ≈ 0.782x and BE: y ≈ -0.726x + 0.726. Their intersection is point H.

Solve for x and y:

0.782x = -0.726x + 0.726

0.782x + 0.726x = 0.726

1.508x ≈ 0.726

x ≈ 0.726 / 1.508 ≈ 0.4816

Then y ≈ 0.782 * 0.4816 ≈ 0.376

Therefore, coordinates of H ≈ (0.4816, 0.376)

Now, we need to find angle AHB. Points A(0, 0), H(0.4816, 0.376), B(1, 0). Let me compute vectors HA and HB, then compute the angle between them.

Vector HA = A - H = (0 - 0.4816, 0 - 0.376) = (-0.4816, -0.376)

Vector HB = B - H = (1 - 0.4816, 0 - 0.376) = (0.5184, -0.376)

Now, angle between vectors HA and HB can be found using the dot product formula:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB = (-0.4816)(0.5184) + (-0.376)(-0.376)

First term: -0.4816 * 0.5184 ≈ -0.250

Second term: 0.376^2 ≈ 0.141

Total ≈ -0.250 + 0.141 ≈ -0.109

Now compute |HA|: sqrt((-0.4816)^2 + (-0.376)^2) ≈ sqrt(0.2319 + 0.1414) ≈ sqrt(0.3733) ≈ 0.611

Compute |HB|: sqrt((0.5184)^2 + (-0.376)^2) ≈ sqrt(0.2688 + 0.1414) ≈ sqrt(0.4102) ≈ 0.6405

Therefore, cos(theta) ≈ -0.109 / (0.611 * 0.6405) ≈ -0.109 / 0.391 ≈ -0.279

Therefore, theta ≈ arccos(-0.279) ≈ 106.3 degrees

Which is approximately 106 degrees, which matches our previous result. Therefore, this numerical calculation confirms that angle AHB is approximately 106 degrees. Therefore, the answer is 106°, so boxed answer is \boxed{106}.

**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 begin by analyzing the given information and applying geometric properties.

---

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

We are given:
- $ \angle BAC = 54^\circ $
- $ \angle ABC = 52^\circ $

Using the triangle angle sum property:
$$
\angle BAC + \angle ABC + \angle ACB = 180^\circ
$$
$$
54^\circ + 52^\circ + \angle ACB = 180^\circ
$$
$$
\angle ACB = 180^\circ - 106^\circ = 74^\circ
$$

---

### Step 2: Use the Property of the Orthocenter

In any triangle, the **orthocenter** $ H $ is the point where the three altitudes intersect. A key geometric property is that the angle formed at the orthocenter by two altitudes (e.g., $ \angle AHB $) is related to the third angle of the triangle.

Specifically, the following identity holds:
$$
\angle AHB = 180^\circ - \angle ACB
$$

This is because the altitudes form right angles with the opposite sides, and the triangle formed by the orthocenter and two vertices has angles that are supplementary to the original triangle's angles.

---

### Step 3: Apply the Formula

We already found:
$$
\angle ACB = 74^\circ
$$

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

---

### 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 $180^\circ - \angle ACB = \angle BAC + \angle ABC = 54^\circ + 52^\circ = \boxed{106^\circ}$.