Chapter 2.4, Problem 20PSC

To determine

To calculate: Measure of $\angle 1$ .

Answer to Problem 20PSC

  ${37}^{\circ }$

Explanation of Solution

Given information: $\angle 1={(x^2+3y)}^{\circ };\angle 2={(20y+3)}^{\circ };\angle 3={(4x+3y)}^{\circ }$

Formula used: Linear pair axiom and vertically opposite angles are equal .

Calculation: $\angle 2+\angle 3={180}^{\circ }$ (linear pair axiom)

  $\begin{array}{l}\therefore 20y+3+4x+3y={180}^{\circ }\\ \therefore 4x+23y={177}^{\circ }\mathrm{....}(1)\end{array}$

  $\angle 1=\angle 3$ (Vertically opposite angles )

  $\begin{array}{l}\therefore x^2+3y=4x+3y\\ \therefore x^2-4x=0\\ \therefore x(x-4)=0\end{array}$

  $\therefore x=0$ or $x=4$

Reject $x=0$

  $\therefore x=4$ And from equation (1), Y=7.

  $\begin{array}{l}\therefore \angle 1=x^2+3y\\ =16+21\\ ={37}^{\circ }\end{array}$