Chapter 15, Problem 21CR

a.

To determine

To find: the length of RS .

Answer to Problem 21CR

  $RS=16$ .

Explanation of Solution

Given:

  

Calculation:

According to the question, there is a circle in which $PT=4,TQ=16,PS=5$ .

The objective is to find RS .

Now, apply secant power theorem,

  $\begin{array}{l}PT\left(PT+PQ\right)=PS\left(PS+RS\right)\\ 4\left(4+16\right)=5\left(5+RS\right)\{\text{substitutes these value}}\\ \text{4}\times \text{20}=5\left(5+RS\right)\\ \left(5+RS\right)=\frac{\text{4}\times \text{20}}{5}=16\\ RS=16-5=11\end{array}$

Therefore, the length of $RS=16$ .

b.

To determine

To find: the angle of $QTS$ .

Answer to Problem 21CR

  $QTS={180}^{\circ }$ .

Explanation of Solution

Given:

  

Calculation:

According to the question, there is a circle in which $PT=4,TQ=16,PS=5$ .

The objective is to find $QTS$ .

Now, apply Pythagoras theorem, to check the triangles are right triangle or not.

  $\begin{array}{l}{12}^2+{\left(11+5\right)}^2={\left(16+4\right)}^2\{\text{subtutes these value}}\\ 144+256=400\left\{\text{simplify}\right\}\\ 400=400\end{array}$

Therefore, $m\angle QRS={90}^{\circ }$ . Now,

  $\begin{array}{l}m\angle QRS=\frac{1}{2}mQTS\left\{\text{inscribed angle theorem}\right\}\\ {90}^{\circ }=\frac{1}{2}mQTS\\ mQTS={180}^{\circ }\end{array}$

Therefore, the angle of $QTS={180}^{\circ }$ .