Chapter 11.1, Problem 23IP

To determine

To find: $m\angle M$ and $m\angle P$

Answer to Problem 23IP

  $\begin{array}{l}m\angle M={27}^0\\ m\angle P={63}^0\end{array}$

Explanation of Solution

Given information:

  $m\angle M={\left(3x-45\right)}^0$

  $m\angle P={\left(2x+15\right)}^0$

  $m\angle M$ and $m\angle P$ are complementary angles.

Given

  $m\angle M$ and $m\angle P$ are complementary angles.

Therefore,

Sum of their measure is ${90}^0$

  $m\angle M+m\angle P={90}^0$

On substituting values

We get

  $\begin{array}{l}m\angle M+m\angle P={90}^0\\ \Rightarrow 3x-{45}^0+2x+{15}^0={90}^0\\ \Rightarrow 5x-{30}^0={90}^0\\ \Rightarrow 5x={90}^0+{30}^0\\ \Rightarrow 5x={120}^0\\ \Rightarrow x=\frac{{120}^0}{5}\\ \Rightarrow x={24}^0\end{array}$

Therefore,

  $m\angle M={\left(3x-45\right)}^0$

On substituting value of $x={24}^0$

We get

  $\begin{array}{l}m\angle M=3x-{45}^0\\ \Rightarrow m\angle M=3\left({24}^0\right)-{45}^0\\ \Rightarrow m\angle M={72}^0-{45}^0\\ \Rightarrow m\angle M={27}^0\end{array}$

  $m\angle P={\left(2x+15\right)}^0$

On substituting value of $x={24}^0$

We get

  $\begin{array}{l}m\angle P=2x+{15}^0\\ \Rightarrow m\angle P=2\left({24}^0\right)+{15}^0\\ \Rightarrow m\angle P={48}^0+{15}^0\\ \Rightarrow m\angle P={63}^0\end{array}$

Therefore,

  $\begin{array}{l}m\angle M={27}^0\\ m\angle P={63}^0\end{array}$