Chapter 7.1, Problem 3PSA

To determine

To calculate: Measure of $\angle C$ and $\angle AFB$

Answer to Problem 3PSA

Measure of $\angle C$ is $40°$ and $\angle AFB$ is $110°$ .

Explanation of Solution

Given information : Here, we have given that $\angle CAB=80°$ , $\angle CBA=60°$ and $\overline{AE}$ and $\overline{BD}$ are altitudes.

Formula used : Sum of all interior angles of triangle is $180°$

Calculation: In $\Delta CAB$ ,

  $\begin{array}{l}\angle ACB+\angle CAB+\angle CBA=180°\\ \Rightarrow \angle ACB+80°+60°=180°\\ \Rightarrow \angle ACB+140°=180°\\ \Rightarrow \angle ACB=180°-140°\\ \Rightarrow \angle ACB=40°\end{array}$

As, $\overline{AE}$ is an altitude,

  $\begin{array}{l}\Rightarrow \angle FAB=\frac{1}{2}\angle CAB\\ \Rightarrow \angle FAB=\frac{1}{2}\times 80°\\ \Rightarrow \angle FAB=40°\end{array}$

Similarly, as $\overline{BD}$ is an altitude,

  $\begin{array}{l}\Rightarrow \angle FBA=\frac{1}{2}\angle CBA\\ \Rightarrow \angle FBA=\frac{1}{2}\times 60°\\ \Rightarrow \angle FBA=30°\end{array}$

Now, in $\Delta FAB$ ,

  $\begin{array}{l}\angle FAB+\angle AFB+\angle FBA=180°\\ \Rightarrow 40°+\angle AFB+30°=180°\\ \Rightarrow 70°+\angle AFB=180°\\ \Rightarrow \angle AFB=180°-70°\\ \Rightarrow \angle AFB=110°\end{array}$