Chapter 10, Problem 23RP

To determine

To find :The value of $x$ and $y$

Answer to Problem 23RP

  $\angle x=6$ and $\angle y=18$

Explanation of Solution

Given information :

We know that, AE=5, AB=10, DP=9, CP=3.

  

Formula Used:

Secant − Tangent Theorem

  $A{B}^2=AE\times AD$

Calculation :

  $\begin{array}{l}\therefore A{B}^2=AE\times AD\Rightarrow {10}^2=5\times (5+\text{x}+9)\\ \Rightarrow \text{x}=6\end{array}$

In $\Delta PEP$ and $\Delta PCD$

  $\angle PEB=\angle PCD$     [Angles in same segment]

  $\angle EPB=\angle CDP$     [Angles in same segment]

And $\angle EPB=\angle CPD$     [Vertically opposite angle]

  $\therefore \Delta PED~\Delta PCD$     [By AAA similarity]

  $\begin{array}{l}\frac{\text{EP}}{\text{CP}}=\frac{\text{BP}}{\text{DP}}\\ \Rightarrow \frac{x}{3}=\frac{y}{9}\\ \Rightarrow y=18\end{array}$