Chapter 11.1, Problem 22IP

To determine

To find: $m\angle TRS$

Answer to Problem 22IP

  $m\angle TRS={30}^0$

Explanation of Solution

Given information:

  $m\angle QRS={158}^0$

  $m\angle TRS=2x^0$

  $m\angle QRT=8x^0+8^0$

From the given figure

We can say that

  $m\angle QRS=m\angle TRS+m\angle QRT$

On substituting values

We get

  $\begin{array}{l}2x^0+8x^0+8^0={158}^0\\ \Rightarrow 10x^0+8^0={158}^0\\ \Rightarrow 10x^0={158}^0-8^0\\ \Rightarrow 10x^0={150}^0\\ \Rightarrow x^0=\frac{{150}^0}{10}\\ \Rightarrow x^0={15}^0\end{array}$

Given

  $m\angle TRS=2x^0$

On substituting values of $x$

We get

  $\begin{array}{l}m\angle TRS=2x^0\\ \Rightarrow m\angle TRS=2\left({15}^0\right)\\ \Rightarrow m\angle TRS={30}^0\end{array}$

Therefore,

  $m\angle TRS={30}^0$