Question

in the diagram below, m∠cih = 107° and m∠bgd = 30°. find m∠fhg. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale. answer attempt 1 out of 2 m∠fhg =

Explanation:

Step1: Identify vertical - angles

$\angle CIH$ and $\angle EIG$ are vertical - angles. So, $m\angle EIG=m\angle CIH = 107^{\circ}$.

Step2: Use the angle - relationship in triangle

In the triangle formed by points $E$, $I$, and $H$, we know that the sum of the interior angles of a triangle is $180^{\circ}$. Also, $\angle EIG$ is an exterior angle of the triangle with angles $\angle EIH$ and $\angle IHG$. We know that $\angle EIG$ and $\angle EIH$ are supplementary. So, $m\angle EIH=180 - 107=73^{\circ}$.
We are given that $\angle BGD = 30^{\circ}$, and $\angle BGD$ and $\angle IHG$ are corresponding angles (assuming parallel lines or relevant angle - relationships based on the context of the problem). So, $m\angle IHG = 30^{\circ}$.

Step3: Calculate $m\angle FHG$

In the straight - line $EF$, $\angle EIH+\angle FHG+\angle IHG = 180^{\circ}$. We know $m\angle EIH = 73^{\circ}$ and $m\angle IHG = 30^{\circ}$. Then $m\angle FHG=180-(73 + 30)=77^{\circ}$.

Answer:

$77$