Chapter 2.4, Problem 19PSB

To determine

To calculate: measure of $\angle QRS$

Answer to Problem 19PSB

  ${98}^{\circ }$

Explanation of Solution

Given information: $\angle PQR$is supplementary with $\angle QRS$ , $\angle QRS$ is supplementary with $\angle TWX,\angle PQR={(5x-48)}^{\circ },\angle TWX={(2x+30)}^{\circ }$ .

Calculation:

  $\begin{array}{l}\angle PQR+\angle QRS={180}^{\circ }\mathrm{....}(1)\\ \angle QRS+\angle TWX={180}^{\circ }\mathrm{....}(2)\end{array}$

Subtracting equation (2) from (1).

  $\begin{array}{l}\angle PQR-\angle TWX=0\\ $ {(5x-48)}^{\circ }-{(2x-30)}^{\circ }=0$3x={78}^{\circ }\\ \therefore x={26}^{\circ }\\ $ \therefore \angle PQR={(5\times 26-48)}^{\circ }$={82}^{\circ }\end{array}$

From equation (1),

  $\begin{array}{l}{82}^{\circ }+\angle QRS={180}^{\circ }\\ \therefore \angle QRS={98}^{\circ }\end{array}$