Chapter 1.3, Problem 15PSC

To determine

To Find: $\text{The values of }m\left(\angle 1\right)\text{,}m\left(\angle 2\right)\text{ and }m\left(\angle 3\right).$ .

Answer to Problem 15PSC

  $\begin{array}{l}m\left(\angle 1\right)=80°\\ m\left(\angle 2\right)=100°\\ m\left(\angle 3\right)=80°\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}m\left(\angle 1\right)=2x+40\\ m\left(\angle 2\right)=2y+40\\ m\left(\angle 3\right)=x+2y\end{array}$

  

Concept Used:

Sum of linear pair of angles is $180°$ and vertically opposite angles are equal.

Calculation:

Here

  $\begin{array}{l}m\left(\angle 1\right)=2x+40°\\ m\left(\angle 2\right)=2y+40°\\ m\left(\angle 3\right)=x+2y\end{array}$

  $\begin{array}{l}\Rightarrow \angle 1+\angle 2=180°\\ \Rightarrow 2x+40°+2y+40°=180°\\ \Rightarrow 2x+2y=100°\\ \Rightarrow x+y=50°\left(1\right)\\ \text{and }\angle 1=\angle 3\\ TRX=90°\\ \Rightarrow 2x+40°=x+2y\\ \Rightarrow x-2y=-40°\left(2\right)\\ \text{Operate }\left(1\right)-\left(2\right),\text{we get}\\ 3y=90°\\ \Rightarrow y=30°\left(3\right)\\ \text{Using}\left(3\right),\text{in }\left(2\right)\text{ get}\\ x=2\times 30°-40°\\ \Rightarrow x=20°\\ \therefore m\left(\angle 1\right)=2\times 20°+40°=80°\\ m\left(\angle 2\right)=2\times 30°+40°=100°\\ m\left(\angle 3\right)=20°+2\times 30°=80°\end{array}$

Conclusion:

  $\begin{array}{l}m\left(\angle 1\right)=80°\\ m\left(\angle 2\right)=100°\\ m\left(\angle 3\right)=80°\end{array}$