Chapter 12.3, Problem 36E

To determine

To show: that the sum of the exterior angle, one exterior angle at each vertex is ${360}^{\circ }$ of quadrilateral where its interior angles are $x^{\circ },y^{\circ },z^{\circ },{\omega }^{\circ }$ .

Answer to Problem 36E

the sum of the exterior angle, one exterior angle at each vertex is ${360}^{\circ }$ of quadrilateral where its interior angles are $x^{\circ },y^{\circ },z^{\circ },{\omega }^{\circ }$ .

Explanation of Solution

Given : the anglesprovided in the question,

  $x^{\circ },y^{\circ },z^{\circ },{\omega }^{\circ }$

Calculation:

the measure of sum exterior angle of each vertex and interior angle = ${180}^{\circ }$

sum of exterior angle at each vertex=

  $\begin{array}{l}{180}^{\circ }-x^{\circ }+{180}^{\circ }-y^{\circ }+{180}^{\circ }-z^{\circ }+{180}^{\circ }-{\omega }^{\circ }\text{ [add angles]}\\ {720}^{\circ }-(x^{\circ }+y^{\circ }+z^{\circ }+{\omega }^{\circ })\text{ [sum of exterior 36}{0}^{\circ }]\\ {720}^{\circ }-{360}^{\circ }\text{ [subtract]}\\ {360}^{\circ }\end{array}$

Hence,the sum of the exterior angle, one exterior angle at each vertex is ${360}^{\circ }$ of quadrilateral where its interior angles are $x^{\circ },y^{\circ },z^{\circ },{\omega }^{\circ }$ .