Chapter 10, Problem 17CUR

To determine

Prove :

  $\Delta OPQ$ is $45°-45°-90°\Delta$

Explanation of Solution

Given:

  $\odot O;m\angle 1=45°$

  

Calculation:

  $\begin{array}{l}m\angle POQ=2\left(m\angle 1\right)\mathrm{...........}[\text{subtended at center by same chord PQ]}\\ \Rightarrow m\angle POQ=2(45°)\\ \Rightarrow m\angle POQ=90°\end{array}$

So, $\Delta OPQ$ is right angled triangle.

Also , in $\Delta OPQ$ , $OP=OQ\mathrm{.............}[radius]$

So, $\Delta OPQ$ is right angle isosceles triangle.

Hence , $\Delta OPQ$ is $45°-45°-90°\Delta$

Hence proved.