Question

1: 'O' is center of circle $C_1$ and AB is the diameter of circle $C_2$. Quadrilateral PQRS is inscribed in the circle with centre 'O'. FInd $\angle QRS$
2. In figure $\angle ADE$ and $\angle ABC$ differ by ${15}^{\circ }$, Find $\angle CAE$

Answer 1

R.E.F image

As angle subtended by the diameter on

any point on the circle is ${90}^{\circ },\angle AOB=90$

(AB diameter in $C_2$ circle)$\Rightarrow \angle SOQ={90}^{\circ }$

Also, angle subtended by any arc on the

center is twice to that subtended

on any other point the circle,hence in

circle $C_1,\angle SOQ=2\times \angle QRS$ (By arc SRQ)

$\Rightarrow \angle QRS=\frac{\angle SOQ}{2}=\frac{90}{2}={45}^{\circ }$

(15)

By same property

mention above,angles

$\angle ADC$ and $\angle ABC$ are equal

(by arc AC)

Now, $\angle ADE-\angle ABC={15}^{\circ }$

$(\angle ADC+\angle CDE)-\angle ABC={15}^{\circ }$

$\Rightarrow (\angle ABC+\angle CDE)-\angle ABC={15}^{\circ }$

$\Rightarrow \angle CED={15}^{\circ }$

And angles subtened by are $CE,\angle CDE=\angle CAE$

$\Rightarrow \angle CAE={15}^{\circ }$