Question

In triangle, F, G, H, comma△FGH, start overline, H, F, end overline, \cong, start overline, G, H, end overline
HF
≅
GH
and m, angle, H, equals, 155, degrees, .m∠H=155
∘
. Find m, angle, G, .m∠G

Answer 1

In triangle \( \triangle FGH \), we are given that segments \( \overline{HF} \) and \( \overline{GH} \) are congruent, which means \( HF \cong GH \). This indicates that triangle \( \triangle FGH \) is isosceles with \( HF \) and \( GH \) as the legs.

Given:
- \( m\angle H = 155^\circ \)

Since triangle \( FGH \) is isosceles, the angles opposite the congruent sides will be equal. Thus, we have:
\[
m\angle F = m\angle G
\]

The sum of the interior angles of a triangle is always \( 180^\circ \). Therefore, we can express this as:
\[
m\angle F + m\angle G + m\angle H = 180^\circ
\]

Substituting the value of \( m\angle H \):
\[
m\angle F + m\angle G + 155^\circ = 180^\circ
\]

Now, since \( m\angle F = m\angle G \), we can let \( m\angle F = m\angle G = x \):
\[
x + x + 155^\circ = 180^\circ
\]
\[
2x + 155^\circ = 180^\circ
\]

Subtracting \( 155^\circ \) from both sides:
\[
2x = 25^\circ
\]

Dividing by \( 2 \):
\[
x = 12.5^\circ
\]

Thus, we find:
\[
m\angle G = 12.5^\circ
\]

The measure of angle \( G \) is:
\[
\boxed{12.5^\circ}
\]