Chapter 10.2, Problem 23E

To determine

To calculate $\angle A,\angle C,\angle D$

Answer to Problem 23E

  $\angle A={88}^{\circ },\angle C={44}^{\circ },\angle D={114}^{\circ }$

Explanation of Solution

Given information:

AB and AD are congruent, BC and DC are congruent and $\angle B={114}^{\circ }$ .

  $\angle A$ is twice

  $\angle C$

Formula used:Properties of kite andangle sum property of quadrilateral,

  $a1+a2+a3+a4={360}^{\circ }$ .

Calculation:

The sides AB and AB are congruent, and BC and DC are congruent. Therefore, the angles between them are congruent, too. Hence,

  $\angle B=\angle D$

Therefore,

  $\angle D={114}^{\circ }$

Let, the $\angle C$ be “

  $x$ ”

Since,

  $\angle A$ is twice

  $\angle C$ ,

  $\angle A$ =2

  $x$

Substituting the values of all the angles in the equation of angle sum property of quadrilateral,

  $\begin{array}{l}2x+114+x+114=360\\ 3x+228=360\\ \end{array}$

Shifting the right-hand side and left-hand side values,

  $\begin{array}{l}3x=360-228\\ 3x=132\end{array}$

Cross-multiplying the values,

  $\begin{array}{l}x=\frac{132}{3}\\ x=44\end{array}$

Hence,

  $\begin{array}{l}\angle C={44}^{\circ }\\ \angle A=2x\\ \angle A=2\times 44\\ \angle A={88}^{\circ }\end{array}$

Therefore,

  $\angle A={88}^{\circ },\angle C={44}^{\circ },\angle D={114}^{\circ }$