Chapter 11.1, Problem 24IP

To determine

To find: Value of $m\angle D$ and $x$

Answer to Problem 24IP

  $\begin{array}{l}x={38.66}^0\\ m\angle D={49.96}^0\end{array}$

Explanation of Solution

Given information:

  $m\angle C={\left(4x-24.6\right)}^0$

  $m\angle D={\left(x+11.3\right)}^0$

  $m\angle C$ and $m\angle D$ are supplementary angles.

Given

  $m\angle C$ and $m\angle D$ are supplementary angles.

Therefore,

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

  $m\angle C+m\angle D={180}^0$

On substituting values

We get

  $\begin{array}{l}m\angle C+m\angle D={180}^0\\ \Rightarrow 4x-{24.6}^0+x+{11.3}^0={180}^0\\ \Rightarrow 5x-{13.3}^0={180}^0\\ \Rightarrow 5x={180}^0+{13.3}^0\\ \Rightarrow 5x={193.3}^0\\ \Rightarrow x=\frac{{193.3}^0}{5}\\ \Rightarrow x={38.66}^0\end{array}$

Therefore,

Value of $x$ is ${38.66}^0$

  $m\angle D={\left(x+11.3\right)}^0$

On substituting value of $x={38.66}^0$

We get

  $\begin{array}{l}m\angle D=x+{11.3}^0\\ \Rightarrow m\angle D={38.66}^0+{11.3}^0\\ \Rightarrow m\angle D={49.96}^0\end{array}$

Therefore,

  $\begin{array}{l}x={38.66}^0\\ m\angle D={49.96}^0\end{array}$