Question

part 2 of 2
find m∠khl and m∠fhl.
m∠khl = 20 °. (type an integer or a decimal.)
m∠fhl = \\(\square\\)° (type an integer or a decimal.)
(image of a geometric figure with angles 5y° and (3y + 8)° and a triangle with side 25 and right angles.)

Explanation:

Step1: Find the value of y

We know that \( m\angle KHL = 20^\circ\) and \( m\angle KHL=5y^\circ\). So we set up the equation \( 5y = 20\). Solving for \( y\), we divide both sides by 5: \( y=\frac{20}{5} = 4\).

Step2: Calculate \( m\angle FHL\)

We know that \( m\angle FHL=(3y + 8)^\circ\). Substitute \( y = 4\) into the expression: \( 3\times4+8=12 + 8=20\)? Wait, no, wait. Wait, maybe the triangles are congruent? Wait, the diagram shows that the two right triangles have equal hypotenuses (since the segments are marked equal) and right angles, so by HL congruence, the triangles are congruent. Therefore, the angles \( \angle KHL\) and \( \angle FHL\) should be equal? Wait, no, wait, the angle \( \angle KHL\) is \( 5y\) and \( \angle FHL\) is \( 3y + 8\). Wait, but if the triangles are congruent, then the angles at H should be equal? Wait, maybe I made a mistake. Wait, let's re - examine.

Wait, the problem says \( m\angle KHL = 20^\circ\) (given as filled in). Wait, maybe the original problem is that the two angles \( 5y\) and \( 3y + 8\) are equal because of the angle bisector or congruent triangles. Wait, if \( 5y=20\), then \( y = 4\). Then \( 3y+8=3\times4 + 8=12 + 8 = 20\)? No, that can't be. Wait, maybe the diagram is such that \( \angle KHL\) and \( \angle FHL\) are complementary? Wait, no, the right angles. Wait, maybe the triangles are congruent, so the angles \( \angle KHL\) and \( \angle FHL\) are equal? Wait, but the user says \( m\angle KHL = 20^\circ\) is already filled. Wait, maybe the question is to find \( m\angle FHL\) given that \( m\angle KHL = 20^\circ\) and the angles \( 5y\) and \( 3y + 8\) are related. Wait, if \( 5y=20\), then \( y = 4\), and then \( 3y + 8=3\times4+8 = 20\)? But that would mean they are equal. Alternatively, maybe there is a mistake in my initial assumption. Wait, maybe the two angles \( \angle KHL\) and \( \angle FHL\) are equal because the triangles are congruent (HL congruence: hypotenuse \( HL\) is common, and the legs are equal as the right triangles have equal - marked sides). So by HL, \( \triangle KHL\cong\triangle FHL\), so \( \angle KHL\cong\angle FHL\). But the user says \( m\angle KHL = 20^\circ\), so \( m\angle FHL = 20^\circ\)? Wait, no, that seems odd. Wait, maybe the original problem was to solve for \( y\) first. Wait, if \( m\angle KHL=5y = 20\), then \( y = 4\). Then \( m\angle FHL=3y + 8=3\times4+8 = 20\). So the measure of \( \angle FHL\) is \( 20^\circ\)?

Wait, maybe I misread the diagram. Let's assume that the two angles \( 5y\) and \( 3y + 8\) are equal (because of congruent triangles). So \( 5y=3y + 8\), then \( 5y-3y=8\), \( 2y = 8\), \( y = 4\). Then \( 5y=20\) and \( 3y + 8=20\). So \( m\angle FHL = 20^\circ\).

Answer:

\( m\angle FHL=\boxed{20}\)