Question

In ΔGHI, start overline, G, I, end overline
GI
is extended through point I to point J, m, angle, I, G, H, equals, left parenthesis, x, plus, 18, right parenthesis, degreesm∠IGH=(x+18)
∘
, m, angle, G, H, I, equals, left parenthesis, 3, x, plus, 8, right parenthesis, degreesm∠GHI=(3x+8)
∘
, and m, angle, H, I, J, equals, left parenthesis, 6, x, plus, 6, right parenthesis, degreesm∠HIJ=(6x+6)
∘
. Find m, angle, I, G, H, .m∠IGH.
Answer

Answer 1

To solve for \( m\angle IGH \) (which is \( (x + 18)^\circ \)), we can use the fact that the angles formed around point \( I \) on line \( GI \) must sum to \( 180^\circ \).

The three angles around point \( I \) are \( m\angle IGH \), \( m\angle GHI \), and \( m\angle HIJ \):
- \( m\angle IGH = x + 18 \)
- \( m\angle GHI = 3x + 8 \)
- \( m\angle HIJ = 6x + 6 \)

We can set up the equation:
\[
m\angle IGH + m\angle GHI + m\angle HIJ = 180^\circ
\]
Substituting in the expressions for the angles:
\[
(x + 18) + (3x + 8) + (6x + 6) = 180
\]

Combining like terms:
\[
x + 3x + 6x + 18 + 8 + 6 = 180
\]
This simplifies to:
\[
10x + 32 = 180
\]

Now, we solve for \( x \):
\[
10x = 180 - 32
\]
\[
10x = 148
\]
\[
x = \frac{148}{10} = 14.8
\]

Now, substituting \( x \) back to find \( m\angle IGH \):
\[
m\angle IGH = x + 18 = 14.8 + 18 = 32.8
\]

Thus,
\[
\boxed{32.8^\circ}
\]
is the measure of \( m\angle IGH \).