Question

find m∠khl and m∠fhl.
m∠khl = \square° (type an integer or a decimal.)
(image shows a diagram with angle khl labeled 5y°, angle fhl labeled (3y + 8)°, and triangle with sides marked equal and right angles.)

Explanation:

Step1: Identify congruent triangles

The triangles \( \triangle KHL \) and \( \triangle FHL \) are congruent (HL congruence, since \( HL = HL \), \( KL = FL \), and right angles). So the angles \( \angle KHL \) and \( \angle FHL \) related to the angle \( 5y^\circ \) and \( (3y + 8)^\circ \) should satisfy \( 5y=3y + 8 \) (since the angles are equal as the triangles are congruent and the segments \( HK \) and \( HF \) form angles with \( HL \) that are equal due to congruence).

Step2: Solve for y

Solve the equation \( 5y=3y + 8 \). Subtract \( 3y \) from both sides: \( 5y-3y=3y + 8-3y \), which gives \( 2y = 8 \). Then divide both sides by 2: \( y=\frac{8}{2}=4 \).

Step3: Find \( m\angle KHL \)

Substitute \( y = 4 \) into \( 5y^\circ \). So \( m\angle KHL=5\times4 = 20^\circ \)? Wait, no, wait. Wait, the angle \( \angle KHL \) is \( 5y \), but wait, maybe I made a mistake. Wait, the triangles are right triangles with \( KL = FL \) (marked congruent), \( HL \) is common, and right angles at \( K \) and \( F \). So \( \triangle KHL\cong\triangle FHL \) by HL. Therefore, \( \angle KHL=\angle FHL \)? Wait, no, the angles at \( H \) are \( 5y \) and \( 3y + 8 \). Wait, maybe the two angles \( 5y \) and \( 3y + 8 \) are equal? Wait, no, maybe the lines \( HK \) and \( HF \) are such that \( HL \) is the angle bisector? Wait, no, the triangles are congruent, so the angles at \( H \) should be equal. So \( 5y=3y + 8 \), solving gives \( y = 4 \), then \( 5y=20 \), but wait, maybe the angle \( \angle KHL \) is \( 5y \), but let's check again. Wait, the diagram shows that \( KL = FL \) (marked with two ticks), \( HL \) is common, and right angles at \( K \) and \( F \), so \( \triangle KHL\cong\triangle FHL \) (HL). Therefore, \( \angle KHL=\angle FHL \), so \( 5y=3y + 8 \), so \( y = 4 \), then \( m\angle KHL=5\times4 = 20 \)? Wait, no, that can't be, because if we have a right triangle, maybe the angle is different. Wait, maybe I misread the diagram. Wait, the angle at \( H \) for \( \triangle KHL \) is \( 5y \), and for \( \triangle FHL \) is \( 3y + 8 \), and since the triangles are congruent, those angles are equal. So solving \( 5y=3y + 8 \) gives \( y = 4 \), so \( 5y=20 \), so \( m\angle KHL = 20^\circ \)? Wait, but let's check again. Wait, maybe the angle \( \angle KHL \) is \( 5y \), and when \( y = 4 \), \( 5\times4 = 20 \). So that's the measure.

Answer:

\( 20 \)