Chapter 11.1, Problem 26IP

To determine

To find: Value of $x,m\angle 1$ and $m\angle 5$

Answer to Problem 26IP

  $\begin{array}{l}m\angle 1={130}^0\\ m\angle 5={50}^0\end{array}$

Value of $x$ is

  $x={26}^0$

Explanation of Solution

Given information:

  $\begin{array}{l}m\angle 1=5x^0\\ m\angle 5={\left(2x-2\right)}^0\end{array}$

From the given figure

We can say that both angles forms a straight line

It means both angles are supplementary angles.

Therefore,

Sum of their measure is ${180}^0$

We get

  $m\angle 1+m\angle 5={90}^0$

On substituting values

We get

  $\begin{array}{l}m\angle 1+m\angle 5={180}^0\\ \Rightarrow 5x^0+2x^0-2^0={180}^0\\ \Rightarrow 5x^0+2x^0={180}^0+2^0\\ \Rightarrow 5x^0+2x^0={182}^0\\ \Rightarrow 7x^0={182}^0\\ \Rightarrow x^0=\frac{{182}^0}{7}\\ \Rightarrow x^0={26}^0\end{array}$

Therefore,

Value of $x$ is ${26}^0$

We know

  $m\angle 1=5x^0$

On substituting value of $x$

We get

  $\begin{array}{l}m\angle 1=5{\left(26\right)}^0\\ m\angle 1={130}^0\end{array}$

We know

  $m\angle 5={\left(2x-2\right)}^0$

On substituting value of $x$

We get

  $\begin{array}{l}m\angle 5={\left(2x-2\right)}^0\\ \Rightarrow m\angle 5=2\left({26}^0\right)-2^0\\ \Rightarrow m\angle 5={52}^0-2^0\\ \Rightarrow m\angle 5={50}^0\end{array}$

Therefore,

  $\begin{array}{l}m\angle 1={130}^0\\ m\angle 5={50}^0\end{array}$