Okay, so I need to find the measure of angle AHB in triangle ABC where the altitudes AD and BE intersect at H. The given angles are angle BAC, which is 54 degrees, and angle ABC, which is 52 degrees. Hmm, let's start by visualizing the triangle and the altitudes.

First, let me recall that in a triangle, the altitude from a vertex is a perpendicular line to the opposite side. So, AD is the altitude from A to BC, and BE is the altitude from B to AC. Their intersection point H is the orthocenter of triangle ABC. The orthocenter is where all three altitudes meet, but since we only have two here, their intersection is H.

Now, the angles at vertices A and B are given. Since the sum of angles in a triangle is 180 degrees, angle ACB must be 180 - 54 - 52 = 74 degrees. So, angle C is 74 degrees. That might be useful later.

I need to find angle AHB. Let me think about the properties related to the orthocenter. The orthocenter has some interesting relationships with the original triangle. For example, the angles at the orthocenter relate to the original angles of the triangle. But I need to figure out exactly how.

Alternatively, maybe I can use coordinate geometry. If I assign coordinates to the triangle's vertices, I can calculate the coordinates of H and then compute the angle AHB. Let me try that approach.

Let me place point A at the origin (0,0) for simplicity. Let's assume side BC is horizontal for easier calculations. Wait, but since AD is the altitude from A to BC, which is perpendicular to BC. If I place A at (0,0), then BC will lie somewhere in the plane. Maybe it's better to place BC on the x-axis? Let's see.

Wait, if I place BC on the x-axis, then the altitude from A would be vertical. Hmm, but if I do that, point A would be somewhere above the x-axis. Alternatively, maybe place point B at (0,0) and point C at (c,0). Then point A would be somewhere in the plane. Let's try that.

Let me denote the coordinates:

Let’s set point B at (0,0). Let’s place point C at (c,0). Point A will be somewhere in the plane. The altitude from A to BC is AD, which is perpendicular to BC. Since BC is on the x-axis, the altitude AD is vertical. Therefore, the x-coordinate of D must be the same as that of A. Wait, but if BC is from (0,0) to (c,0), then the altitude from A to BC would have to be a vertical line if BC is horizontal. So, if A is at (d, e), then D would be (d,0). But then BE is the altitude from B to AC. Let me try to assign coordinates step by step.

Alternatively, maybe using trigonometry. Since we know angles at A and B, perhaps we can find the sides using the sine law and then compute coordinates.

But this might get complicated. Maybe there's a more straightforward geometric approach.

Wait, another idea: in triangle ABC, the orthocenter H. The angles at H can be related to the original angles. For example, angle AHB is equal to 180 degrees minus angle ACB. Is that a property? Let me recall.

Yes, actually, in the orthocenter configuration, the angles at the orthocenter relate to the original angles. Specifically, angle AHB is equal to 180 degrees minus angle ACB. Let me verify that.

Because the orthocenter H creates several cyclic quadrilaterals. For instance, quadrilateral HBEC is cyclic because angles HBE and HCE are both right angles. Wait, maybe not. Let me think.

Alternatively, consider that angles at H. Since AD and BE are altitudes, they are perpendicular to BC and AC respectively. Therefore, angles HDB and HEB are right angles. But I need to connect this to angle AHB.

Alternatively, maybe using the fact that in triangle ABC, the orthocenter H creates several right angles. For example, triangles AHB, BHC, and AHC are all related to the original angles through their orthocenter.

Wait, another approach: in triangle ABC, the orthocenter H. Then, angle AHB is equal to 180 degrees minus angle ACB. Let me check with a simple case. Suppose triangle ABC is acute. Then, the orthocenter lies inside the triangle. Let's take an equilateral triangle, all angles 60 degrees. Then angle AHB would be 120 degrees, which is 180 - 60. So that seems to hold. If angle ACB is 74 degrees, then angle AHB would be 180 - 74 = 106 degrees. Is that correct?

Wait, let me verify with another example. Suppose triangle ABC has angles 90, 45, 45. Then angle ACB is 45 degrees. Then angle AHB would be 180 - 45 = 135 degrees. But in a right-angled triangle, the orthocenter is at the right-angled vertex, so H would be at point C. Then angle AHB would be angle ACB, which is 45 degrees. Wait, that contradicts. So maybe my initial assumption is wrong.

Wait, no. If triangle ABC is right-angled at C, then the orthocenter is at C. Then angle AHB would be angle ACB, which is 90 degrees. Wait, but in that case, angle ACB is 90 degrees, so 180 - 90 = 90. Which matches. Wait, in the 45-45-90 triangle, angle ACB is 90 degrees, so angle AHB would be 180 - 90 = 90, which is correct. But if angle ACB is 45 degrees, then angle AHB would be 135 degrees. But in a 45-45-90 triangle, angle AHB would be 90 degrees. Wait, maybe my previous reasoning is flawed.

Wait, no. Let me think again. If triangle ABC is right-angled at C, then H is at C. So angle AHB is angle ACB which is 90 degrees. So angle AHB = 90 degrees, which is 180 - 90 = 90. So that works. If angle ACB is 45, then angle AHB is 135. Let's take another triangle where angle ACB is 30 degrees. Then angle AHB would be 150 degrees. Let's see.

Suppose triangle ABC with angle at C is 30 degrees. Let me construct coordinates. Let’s place point C at (0,0), point B at (1,0), and point A somewhere. Let’s compute coordinates.

Wait, maybe this is getting too complicated. Let me check a resource or recall the exact theorem.

Upon recalling, the measure of angle AHB is equal to 180 degrees minus the measure of angle ACB. So, angle AHB = 180° - angle ACB. Therefore, in this problem, angle ACB is 74°, so angle AHB would be 180 - 74 = 106°. Is that correct?

Wait, but in the right-angled triangle example, angle ACB is 90°, so angle AHB would be 90°, which matches. In the equilateral triangle, angle ACB is 60°, angle AHB is 120°, which is 180 - 60. So that works. In the 45-45-90 triangle, angle ACB is 90°, angle AHB is 90°, which is 180 - 90. Wait, but 180 - 90 is 90, which is correct. So maybe the formula holds.

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

Wait, but let me verify with another approach to be sure.

Alternatively, using the properties of cyclic quadrilaterals. Since H is the orthocenter, quadrilaterals like BHEC, AHEB, and AHD etc., are cyclic. Wait, actually, the reflection properties. Let me recall that in the orthocenter configuration, angles subtended by the same segment are equal.

Alternatively, consider triangle AHB. The angles at A and B in triangle AHB can be related to the original triangle.

Wait, in triangle ABC, the orthocenter H. Then, angle AHB is equal to 180° minus angle ACB. So, angle AHB = 180° - angle C. Since angle C is 74°, angle AHB is 106°. Therefore, the answer is 106 degrees.

But let me check with coordinates to be absolutely sure.

Let me assign coordinates to the triangle. Let’s place point A at (0,0), point B at (b,0), and point C at (c,d). Then, compute the coordinates of H by finding the intersection of altitudes AD and BE.

Given angles at A and B: angle BAC = 54°, angle ABC = 52°, so angle ACB = 74°.

First, let's compute the lengths using the Law of Sines.

In triangle ABC, sides opposite angles A, B, C are a, b, c respectively.

Wait, standard notation: side a is opposite angle A, side b opposite angle B, side c opposite angle C.

But here, angle A is 54°, angle B is 52°, angle C is 74°. So:

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

Let’s let’s set the triangle such that side BC is length a, AC is length b, and AB is length c.

But perhaps it's better to assign coordinates. Let me place point A at (0,0), point B at (c,0), and point C at (d,e). Then, angles at A and B are given.

But maybe a better approach is to use coordinate system with point A at (0,0), point B at (1,0), and point C somewhere in the plane. Then compute coordinates of H.

Alternatively, use trigonometric relations.

Let me try this:

Since angle BAC is 54°, angle ABC is 52°, angle ACB is 74°, as established.

Let’s denote triangle ABC with coordinates:

- Let’s place point A at (0,0).

- Let’s place point B at (c,0), so AB is along the x-axis.

- Point C will be somewhere in the plane. Let’s compute coordinates of C.

Using the Law of Sines:

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

But here, side a is BC, side b is AC, side c is AB.

Given angles:

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

Let’s denote the sides:

a = BC = k * sin A = k * sin 54°

b = AC = k * sin B = k * sin 52°

c = AB = k * sin C = k * sin 74°

For some scaling factor k. Let’s choose k = 1 for simplicity. Then:

a = sin 54° ≈ 0.8090

b = sin 52° ≈ 0.7880

c = sin 74° ≈ 0.9613

So, coordinates:

- A: (0,0)

- B: (c,0) ≈ (0.9613, 0)

- C: Let's compute coordinates. Since AC = b ≈ 0.7880, and angle at A is 54°, so coordinates of C can be found as:

From point A at (0,0), moving at an angle of 54° with length AC = 0.7880:

C_x = AC * cos(54°) ≈ 0.7880 * cos(54°) ≈ 0.7880 * 0.5878 ≈ 0.462

C_y = AC * sin(54°) ≈ 0.7880 * 0.8090 ≈ 0.638

So, C is approximately at (0.462, 0.638)

But also, the distance from B to C should be a ≈ 0.8090.

Let me check distance between B (0.9613, 0) and C (0.462, 0.638):

Distance = sqrt[(0.9613 - 0.462)^2 + (0 - 0.638)^2] ≈ sqrt[(0.4993)^2 + (-0.638)^2] ≈ sqrt[0.2493 + 0.4070] ≈ sqrt[0.6563] ≈ 0.810, which is approximately 0.8090. So that checks out.

Now, having coordinates of A, B, C, we can compute the equations of the altitudes AD and BE, find their intersection H, then compute angle AHB.

First, find the equation of altitude AD from A to BC.

Since BC is from (0.9613,0) to (0.462,0.638). The slope of BC is (0.638 - 0)/(0.462 - 0.9613) ≈ 0.638 / (-0.4993) ≈ -1.278.

Therefore, the altitude from A to BC is perpendicular to BC, so its slope is the negative reciprocal of -1.278, which is 1/1.278 ≈ 0.782.

But wait, since BC has slope m, the altitude has slope -1/m. Wait, the slope of BC is (0.638 - 0)/(0.462 - 0.9613) = 0.638 / (-0.4993) ≈ -1.278. Therefore, the altitude from A is perpendicular, so slope is 1/1.278 ≈ 0.782. Wait, but since the slope of BC is negative, the negative reciprocal would be positive. So, yes, slope ≈ 0.782.

But since the altitude passes through A (0,0), its equation is y = 0.782x.

Now, equation of altitude BE from B to AC.

First, find the slope of AC. AC is from A (0,0) to C (0.462,0.638). Slope is (0.638 - 0)/(0.462 - 0) ≈ 0.638 / 0.462 ≈ 1.381.

Therefore, the altitude BE is perpendicular to AC, so its slope is -1/1.381 ≈ -0.724.

Altitude BE passes through point B (0.9613,0). So, equation is y - 0 = -0.724(x - 0.9613), which simplifies to y = -0.724x + 0.724*0.9613 ≈ -0.724x + 0.696.

Now, find intersection point H of the two altitudes AD and BE.

Set equations equal:

0.782x = -0.724x + 0.696

0.782x + 0.724x = 0.696

1.506x = 0.696

x ≈ 0.696 / 1.506 ≈ 0.462

Then y ≈ 0.782 * 0.462 ≈ 0.361

So, coordinates of H are approximately (0.462, 0.361)

Wait, but point C is at (0.462, 0.638). So H is vertically below C on the same x-coordinate. That makes sense since AD is the altitude from A to BC, which is vertical in this coordinate system. Wait, no, in this coordinate system, the altitude from A is not vertical. Wait, hold on.

Wait, in our coordinate system, BC is not vertical. Wait, BC goes from (0.9613,0) to (0.462,0.638). So, it's a diagonal line. The altitude from A to BC is a line with slope ≈0.782, which is not vertical. So, the x-coordinate of H is 0.462, which is the same as point C. Wait, no, point C is at (0.462,0.638). So H is at (0.462, 0.361), which is directly below C on the x=0.462 line. That makes sense because the altitude from A is vertical? Wait, no, because in this coordinate system, BC is not vertical. Wait, perhaps there's a mistake here.

Wait, if the altitude from A is perpendicular to BC, and BC has slope ≈-1.278, then the altitude from A should have slope ≈0.782, which is not vertical. So, the altitude from A is not vertical. Then, why is the x-coordinate of H equal to the x-coordinate of C? That seems contradictory.

Wait, perhaps an error in calculation. Let me check the equations again.

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

Equation of altitude BE: passes through B (0.9613,0) with slope ≈-0.724. So, equation y = -0.724x + 0.724*0.9613.

Calculating 0.724*0.9613: 0.724*0.96 ≈ 0.724*0.96 = 0.69504, plus 0.724*0.0013 ≈ 0.0009412, so total ≈0.69598. So equation is y ≈ -0.724x + 0.696.

Setting equal: 0.782x = -0.724x + 0.696

Adding 0.724x to both sides: 1.506x = 0.696

x ≈ 0.696 / 1.506 ≈ 0.462

Then y ≈0.782*0.462 ≈0.361

So, H is at (0.462, 0.361). But point C is at (0.462,0.638). So H is directly below C on the vertical line x=0.462. So, the altitude from C to AB is vertical? Wait, in this coordinate system, AB is along the x-axis from (0,0) to (0.9613,0). So, the altitude from C to AB would be vertical? Wait, if AB is horizontal, then the altitude from C to AB is vertical. But H is the orthocenter, so all three altitudes meet there.

Wait, but in this coordinate system, the altitude from B to AC is not vertical. Wait, maybe I need to check the altitude from C to AB. Since AB is horizontal, the altitude from C to AB is vertical, dropping straight down from C to AB. So, the foot of the altitude from C is at (0.462,0), which is point D. Then, the altitude from C is the vertical line x=0.462. But H is at (0.462, 0.361), which is on this vertical line. Therefore, H lies on the altitude from C to AB, as it should. Similarly, H lies on altitude from A (AD) and altitude from B (BE). So, that seems consistent.

Therefore, H is at (0.462,0.361). Now, to compute angle AHB. Points A (0,0), H (0.462,0.361), B (0.9613,0). So, angle at H between points A and B.

To compute angle AHB, we can use vectors. The vectors HA and HB. HA is from H to A: (0 - 0.462, 0 - 0.361) = (-0.462, -0.361). HB is from H to B: (0.9613 - 0.462, 0 - 0.361) = (0.4993, -0.361).

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

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB: (-0.462)(0.4993) + (-0.361)(-0.361) ≈ (-0.2305) + (0.1303) ≈ -0.0992

|HA| = sqrt((-0.462)^2 + (-0.361)^2) ≈ sqrt(0.213 + 0.130) ≈ sqrt(0.343) ≈ 0.586

|HB| = sqrt((0.4993)^2 + (-0.361)^2) ≈ sqrt(0.2493 + 0.1303) ≈ sqrt(0.3796) ≈ 0.616

Therefore, cos(theta) ≈ -0.0992 / (0.586 * 0.616) ≈ -0.0992 / 0.362 ≈ -0.274

Therefore, theta ≈ arccos(-0.274) ≈ 106 degrees. Which matches the earlier result.

Therefore, angle AHB is 106 degrees. So, the answer is 106 degrees, which is 180 - 74, where 74 degrees is angle ACB.

Thus, the initial approach was correct. Therefore, angle AHB is 106 degrees.
Therefore, $\angle AHB$ is $\boxed{106^\circ}$.
</think>