Question

In the figure, what is the value of $\angle {\text{PQR}} + \angle {\text{PRQ}}$.
$
  {\text{A}}{\text{. 7}}{{\text{0}}^\circ} \\
  {\text{B}}{\text{. 8}}{{\text{0}}^\circ} \\
  {\text{C}}{\text{. 9}}{{\text{0}}^\circ} \\
  {\text{D}}{\text{. 10}}{{\text{0}}^\circ} \\
 $

Answer 1

Hint: Here, we will proceed by using the property that the angle formed by drawing lines from the ends of the diameter of a circle to its circumference forms a right angle (${\text{9}}{{\text{0}}^\circ}$) and then, using the property of the triangle i.e., the sum of all the interior angles of any triangle is always equal to ${\text{18}}{{\text{0}}^\circ}$.

Complete step-by-step answer:
In the figure, point O is the centre of the circle so the line segment QR is the diameter of the circle because the diameter of any circle passes through the centre of that circle.
As we know that the angle formed by drawing lines from the ends of the diameter of a circle to its circumference forms a right angle (${\text{9}}{{\text{0}}^\circ}$)
Using the above concept, we can say that $\angle {\text{QPR}} = {90^\circ}$
Also we know that according to the property of the triangle, the sum of all the interior angles of any triangle is always equal to ${\text{18}}{{\text{0}}^\circ}$
Using the above property of the triangle, we can write
In $\vartriangle $PQR, $\angle {\text{QPR}} + \angle {\text{PQR}} + \angle {\text{PRQ}} = {180^\circ}{\text{ }} \to {\text{(1)}}$
By putting the value of $\angle {\text{QPR}} = {90^\circ}$ in equation (1), we get
\[
   \Rightarrow {90^\circ} + \angle {\text{PQR}} + \angle {\text{PRQ}} = {180^\circ} \\
   \Rightarrow \angle {\text{PQR}} + \angle {\text{PRQ}} = {180^\circ} - {90^\circ} \\
   \Rightarrow \angle {\text{PQR}} + \angle {\text{PRQ}} = {90^\circ} \\
 \]
Therefore, the value of $\angle {\text{PQR}} + \angle {\text{PRQ}}$ is equal to ${\text{9}}{{\text{0}}^\circ}$
Hence, option C is correct.

Note: In this particular problem, we had to find the value for $\angle {\text{PQR}} + \angle {\text{PRQ}}$ which can be done by forming an equation which will consist of both of these angles (i.e., $\angle {\text{PQR}}$ and $\angle {\text{PRQ}}$). Also, the length OP in the figure will be equal to the radius of the given circle because point O is the centre of the circle and point P is any point on the circumference of the given circle.