In DeltaGHI, bar(GI) is extended through point I to point J, m/_IGH=(x-2)^(@), m/_HIJ=(4x-18)^(@), and m/_GHI=(x+14)^(@). Find m/_HIJ. Answer:

Identify Relationship: Identify the relationship between the angles. Since GI is extended through I to J, angle IGH and angle HIJ form a linear pair and their measures add up to 180 degrees.Set Up Equation: Set up the equation using the angle addition postulate. m/_IGH + m/_HIJ = 180^(@) Substitute the given expressions for m/_IGH and m/_HIJ. (x - 2)^(@) + (4x - 18)^(@) = 180^(@)Combine and Solve: Combine like terms and solve for x. x - 2 + 4x - 18 = 180 5x - 20 = 180 5x = 200 x = 40Substitute Value: Substitute the value of x back into the expression for m/_HIJ. m/_HIJ = (4x - 18)^(@) m/_HIJ = (4(40) - 18)^(@) m/_HIJ = (160 - 18)^(@) m/_HIJ = 142^(@)

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In
DeltaGHI, bar(GI) is extended through point I to point J,
m/_IGH=(x-2)^(@),
m/_HIJ=(4x-18)^(@), and
m/_GHI=(x+14)^(@). Find
m/_HIJ.
Answer:

In $\Delta \mathrm{GHI}, \overline{G I}$ is extended through point I to point J, $\mathrm{m} \angle I G H=(x-2)^{\circ}$ , $\mathrm{m} \angle H I J=(4 x-18)^{\circ}$ , and $\mathrm{m} \angle G H I=(x+14)^{\circ}$ . Find $\mathrm{m} \angle H I J$ . $\newline$ Answer:

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Q. In

\Delta \mathrm{GHI}, \overline{G I}

is extended through point I to point J,

\mathrm{m} \angle I G H=(x-2)^{\circ}

\mathrm{m} \angle H I J=(4 x-18)^{\circ}

, and

\mathrm{m} \angle G H I=(x+14)^{\circ}

. Find

\mathrm{m} \angle H I J

\newline

Answer:

Identify Relationship: Identify the relationship between the angles. $\newline$ Since $GI$ is extended through $I$ to $J$ , angle $IGH$ and angle $HIJ$ form a linear pair and their measures add up to $180$ degrees.
Set Up Equation: Set up the equation using the angle addition postulate. $\newline$ $m/\_IGH + m/\_HIJ = 180^\circ$ $\newline$ Substitute the given expressions for $m/\_IGH$ and $m/\_HIJ$ . $\newline$ $(x - 2)^\circ + (4x - 18)^\circ = 180^\circ$
Combine and Solve: Combine like terms and solve for $x$ . $x - 2 + 4x - 18 = 180$ $5x - 20 = 180$ $5x = 200$ $x = 40$
Substitute Value: Substitute the value of $x$ back into the expression for $m/\_HIJ$ .
$m/\_HIJ = (4x - 18)^{@}$
$m/\_HIJ = (4(40) - 18)^{@}$
$m/\_HIJ = (160 - 18)^{@}$
$m/\_HIJ = 142^{@}$