Chapter 3.4, Problem 24WE

(a)

To determine

To Find: The measure of $\angle FIG$ .

Answer to Problem 24WE

  $m\angle FIG=130°$

Explanation of Solution

Given Information: In the figure shown below,the bisectors of $\angle EFG$ and $\angle EGF$ meet at $I$ .

  

Also, $m\angle EFG=40$ .

Calculation:

Since, $IF$ is the bisector of $\angle EFG$ and $IG$ is the bisector of $\angle EGF$ , $m\angle IGF=\frac{1}{2}m\angle EGF\text{ and }m\angle IFG=\frac{1}{2}m\angle EFG$ .

In $\Delta EGF$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle EGF+m\angle EFG+m\angle FEG=180°\\ m\angle EGF+40°+80°=180°\\ m\angle EGF=180°-120°\\ m\angle EGF=60°\end{array}$

In $\Delta FIG$ ,

  $\begin{array}{l}m\angle FIG+m\angle IGF+m\angle IFG=180°\\ m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180°\\ m\angle FIG+\frac{1}{2}\times 60°+\frac{1}{2}\times 40°=180°\\ m\angle FIG+30°+20°=180°\\ m\angle FIG=180°-50°\\ m\angle FIG=130°\end{array}$

Therefore, the measure of $\angle FIG$ is $130°$ .

(b)

To determine

To Find: The measure of $\angle FIG$ .

Answer to Problem 24WE

  $m\angle FIG=130°$

Explanation of Solution

Given Information: In the figure shown below,the bisectors of $\angle EFG$ and $\angle EGF$ meet at $I$

  

Also, $m\angle EFG=50°$ .

Calculation:

Since, $IF$ is the bisector of $\angle EFG$ and $IG$ is the bisector of $\angle EGF$ , $m\angle IGF=\frac{1}{2}m\angle EGF\text{ and }m\angle IFG=\frac{1}{2}m\angle EFG$

In $\Delta EGF$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle EGF+m\angle EFG+m\angle FEG=180°\\ m\angle EGF+50+80=180°\\ m\angle EGF=180-130°\\ m\angle EGF=50°\end{array}$

In $\Delta FIG$ ,

  $\begin{array}{l}m\angle FIG+m\angle IGF+m\angle IFG=180°\\ m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180°\\ m\angle FIG+\frac{1}{2}\times 50°+\frac{1}{2}\times 50°=180°\\ m\angle FIG+25°+25°=180°\\ m\angle FIG=180°-50°\\ m\angle FIG=130°\end{array}$

Therefore, the measure of $\angle FIG$ is $130°$ .

(c)

To determine

To Generalise: The measure of $\angle FIG$ is $130°$ when $m\angle EFG=40°$ or when $m\angle EFG=50°$ .

Explanation of Solution

Given Information: In the figure shown below,the bisectors of $\angle EFG$ and $\angle EGF$ meet at $I$

  

Calculation:

Since, $IF$ is the bisector of $\angle EFG$ and $IG$ is the bisector of $\angle EGF$ , $m\angle IGF=\frac{1}{2}m\angle EGF\text{ and }m\angle IFG=\frac{1}{2}m\angle EFG$

In $\Delta FIG$ ,

  $\begin{array}{l}m\angle FIG+m\angle IGF+m\angle IFG=180°\\ m\angle FIG+\frac{1}{2}m\angle EGF+\frac{1}{2}m\angle EFG=180°\\ m\angle FIG+\frac{1}{2}\left(m\angle EGF+m\angle EFG\right)=180°\\ m\angle FIG+\frac{1}{2}\left(180°-\angle FEG\right)=180°\\ m\angle FIG+90°-\frac{1}{2}\angle FEG=180°\\ m\angle FIG=90°+\frac{1}{2}\angle FEG\\ m\angle FIG=90°+\frac{1}{2}\times 80°\\ m\angle FIG=90°+40°\\ m\angle FIG=130°\end{array}$

Hence, $m\angle FIG$ is independent of $m\angle EFG$ and is always equal to $130°$ .