# Please solve q 64 ?

May 11, 2018

$\angle Q R P = {55}^{\circ}$

#### Explanation:

Given that, $P R$ is the diameter of circle and $\angle R P S , \angle Q P R , \angle Q R P , \mathmr{and} \angle P R S$ form an $A P$. Also, $\angle R P S = {15}^{\circ}$

Let $\angle Q P R = x \mathmr{and} \angle P R S = y$.

In $\Delta P R S , \angle P R S + \angle P S R + \angle P R S = 180$

$\rightarrow {15}^{\circ} + \angle P R S + {90}^{\circ} = {180}^{\circ}$

$\rightarrow \angle P R S = {75}^{\circ}$

If three numbers $a , b , c$ are in $A P$ then $a + c = 2 b$

${15}^{\circ} , x , y$ and $x , y , {75}^{\circ}$ are in $A P$ as ${15}^{\circ} , x , y , {75}^{\circ}$ are in AP .

So, ${15}^{\circ} + y = 2 x$.....[1]

and $x + {75}^{\circ} = 2 y$.....[2]

From [1], $x = \frac{{15}^{\circ} + y}{2}$

Putting the value of $x$ in eqn [2],

$\rightarrow \frac{15 + {y}^{\circ}}{2} + {75}^{\circ} = 2 y$

$\rightarrow \frac{{15}^{\circ} + y + {150}^{\circ}}{2} = 2 y$

$\rightarrow {165}^{\circ} + y = 4 y$

$\rightarrow y = \angle Q R P = {55}^{\circ}$

So, the correct option is (1).