Question

if \\(\overline{ef} \cong \overline{eh}\\), \\(m\angle egf = s + 8^\circ\\), and \\(m\angle egh = 5s\\), what is the value of \\(s\\)?

Explanation:

Step1: Identify Triangle Congruence

Since \( \overline{EF} \cong \overline{EH} \), \( \angle EFG = \angle EHG = 90^\circ \) (right angles), and \( \overline{EG} \) is common, triangles \( EFG \) and \( EHG \) are congruent (HL congruence). Thus, \( \angle EGF = \angle EGH \)? Wait, no—wait, the angles: \( m\angle EGH = 5s \) and \( m\angle EGF = s + 8 \). Wait, actually, because \( \overline{EF} \cong \overline{EH} \) and \( EG \) is common, the triangles \( EFG \) and \( EHG \) are congruent, so their corresponding angles \( \angle EGF \) and \( \angle EGH \)? Wait, no, maybe \( \angle EGF \) and \( \angle EGH \) are related? Wait, no, the problem says \( m\angle EGF = s + 8^\circ \) and \( m\angle EGH = 5s \). Wait, maybe \( \angle EGF = \angle EGH \)? Wait, no, that can't be. Wait, maybe I misread. Wait, the diagram: \( E \) is the vertex, \( F \) and \( H \) are right angles, so \( EF \perp GF \), \( EH \perp GH \). Since \( EF \cong EH \) and \( EG \) is common, triangles \( EFG \) and \( EHG \) are congruent (HL), so \( \angle EGF \cong \angle EGH \)? Wait, no, \( \angle EGF \) is at \( G \), between \( EG \) and \( FG \); \( \angle EGH \) is at \( G \), between \( EG \) and \( HG \). So if the triangles are congruent, then \( \angle EGF = \angle EGH \)? Wait, but the problem states \( m\angle EGF = s + 8 \) and \( m\angle EGH = 5s \). Wait, that would mean \( s + 8 = 5s \)? Wait, no, maybe I got the angles reversed. Wait, maybe \( \angle EGH = \angle EGF \)? Wait, let's set up the equation. If the triangles are congruent, then \( \angle EGF = \angle EGH \). So \( s + 8 = 5s \)? Wait, no, solving \( s + 8 = 5s \) gives \( 8 = 4s \), so \( s = 2 \)? Wait, no, that seems too small. Wait, maybe I made a mistake. Wait, the problem says \( m\angle EGH = 5s \) and \( m\angle EGF = s + 8 \). Wait, maybe \( \angle EGH = \angle EGF \)? Wait, let's check the problem again: "If \( \overline{EF} \cong \overline{EH} \), \( m\angle EGF = s + 8^\circ \), and \( m\angle EGH = 5s \), what is the value of \( s \)?" So since \( EF \cong EH \), \( \angle EFG = \angle EHG = 90^\circ \), and \( EG \) is common, by HL, \( \triangle EFG \cong \triangle EHG \). Therefore, their corresponding angles \( \angle EGF \) and \( \angle EGH \) are congruent. So \( m\angle EGF = m\angle EGH \). Therefore, \( s + 8 = 5s \). Wait, solving that: \( 8 = 5s - s \) → \( 8 = 4s \) → \( s = 2 \). Wait, but that seems too simple. Wait, maybe I misread the angles. Wait, maybe \( \angle EGH = 5s \) and \( \angle EGF = s + 8 \), but maybe \( \angle EGH = \angle EGF \)? Wait, let's do the math. \( s + 8 = 5s \) → \( 4s = 8 \) → \( s = 2 \). Wait, but let's check. If \( s = 2 \), then \( m\angle EGF = 2 + 8 = 10^\circ \), \( m\angle EGH = 5*2 = 10^\circ \). That works. So the equation is \( s + 8 = 5s \).

Step1: Set up the equation

Since \( \overline{EF} \cong \overline{EH} \) and \( EG \) is common, \( \triangle EFG \cong \triangle EHG \) (HL). Thus, \( \angle EGF \cong \angle EGH \), so \( m\angle EGF = m\angle EGH \).
Given \( m\angle EGF = s + 8^\circ \) and \( m\angle EGH = 5s \), we set:
\( s + 8 = 5s \)

Step2: Solve for \( s \)

Subtract \( s \) from both sides:
\( 8 = 4s \)
Divide both sides by 4:
\( s = \frac{8}{4} = 2 \)

Answer:

\( \boxed{2} \)