Chapter 12.5, Problem 25HP

To determine

To calculate: the value of $x$

Answer to Problem 25HP

The value of $x$ is ${32.5}^{\circ }$

Explanation of Solution

Given information:

System equation: $RT=TS$ and $QR=QT.$

  

Calculation:

Consider the triangle QSR

  

Use the fact that the sum of the angles in a triangle is ${180}^{\circ },$ So,

  $\angle RQS+\angle QSR+\angle SRQ={180}^{\ast }\dots \mathrm{...}(1)$

Consider the triangle RTS. It is known that angles opposite to congruent sides are equal. Use this fact and $\text{RT}=\text{TS},$ to find the angle $\angle SRT$ So,

  $\angle TSR=\angle SRT=x^{\ast }\dots \dots (2)$

Consider the triangle QTR.

Similarly, use $\text{QR}=\text{QT},$ to find the angle, $\angle QRT.\text{So},$

Let,

  $\angle QTR=\angle QRT=y^{\ast }$

Use the fact that the sum of the angles in a triangle is ${180}^{\circ }.$

  $\angle QTR+\angle TRQ+\angle RQT={180}^{\ast }\dots \mathrm{...}(3)$

  $\text{Substitute }{y}^{\ast }\text{ for}\angle QTR,y^{\ast }\text{ for}\angle TRQ,\text{and}{50}^{\ast }\text{ for}\angle RQT,\text{in equation}(3).\text{ So,}$

  $\begin{array}{l}{y}^{\ast }+y^{\ast }+50={180}^{\circ }\\ 2y^{\circ }+{50}^{\circ }-{50}^{\circ }=180-{50}^{\circ }\\ 2y^{\circ }={130}^{\circ }\\ y^{\circ }=\mathrm{65........}\left(4\right)\end{array}$

Use the equation (2), equation (3) and equation (4) to find $\angle SRQ.$ So,

  $\begin{array}{cc}\angle SRQ& =\angle SRT+\angle QRT\\ & =x^2+y^{\ast }\\ & =x^2+{65}^{\circ }\end{array}$

Substitute ${50}^{\circ }$ for $\angle RQS,x^{\ast }$ for $\angle QSR,$ and $x^{\ast }+{65}^{\circ }$ for $\angle SRQ,$ in equation (1) to find the angle, $\angle QSR.$ So,

Therefore, the value of $x$ is ${32.5}^{\circ }$