Chapter 3.4, Problem 34WE

To determine

To find out about the angles of ABCD.

Explanation of Solution

Given:

The given figure is-

  

Calculation:

It is known that the same side angles are supplementary.

  $\begin{array}{l}3\angle x+3\angle y=180\\ 3\left(\angle x+\angle y\right)=180\\ \angle x+\angle y={60}^{\circ }\mathrm{.......}(1)\end{array}$

It is known that the sum of angles in triangle are ${180}^{\circ }$

  $\begin{array}{l}\angle x+\angle y+\angle D=180\\ \angle D=180-{60}^{\circ }={120}^{\circ }\mathrm{.......}(2)\end{array}$

Similarly,

  $\begin{array}{l}2\angle x+2\angle y+\angle B=180\\ 2\left(\angle x+\angle y\right)+\angle B=180\\ 120+\angle B={180}^{\circ }\\ \angle B={60}^{\circ }\mathrm{.......}(3)\end{array}$

Similarly,

  $\begin{array}{l}\angle x+2\angle y+\angle A=180\\ \angle y+\left(\angle x+\angle y\right)+\angle A=180\\ \angle y+\angle A={180}^{\circ }-{60}^{\circ }\\ \angle A={120}^{\circ }-\angle y\mathrm{.......}(3)\end{array}$

Similarly,

  $\begin{array}{l}2\angle x+\angle y+\angle C=180\\ \angle x+\left(\angle x+\angle y\right)+\angle C=180\\ \angle x+\angle C={180}^{\circ }-{60}^{\circ }\\ \angle C={120}^{\circ }-\angle y\mathrm{.......}(3)\end{array}$