Question

In the figure, PQ is the diameter of the circle. If $\angle PQR={{65}^{\circ }}$ , $\angle RPS={{25}^{\circ }}$ and $\angle QPT={{60}^{\circ }}$ , then find the measure of (i) $\angle QPR$ , (ii) $\angle PRS$ , (iii) $\angle PSR$ and (iv) $\angle PQT$ ?

Answer 1

We start solving the problem by making use of the fact that angle in a semicircle is ${{90}^{\circ }}$ to find the value of angle $\angle QTP$ . We then make use of the fact that the sum of the angles in a triangle is ${{180}^{\circ }}$ for triangle PQT to find the angle $\angle PQT$ . We then make use of the fact that angle in a semicircle is ${{90}^{\circ }}$ to find the value of angle $\angle QRP$ . We then make use of the fact that the sum of the angles in a triangle is ${{180}^{\circ }}$ for triangle PQR to find the angle $\angle QPR$ . We then make use of the fact that the sum of opposite angles in a concyclic polynomial is ${{180}^{\circ }}$ to find the angle $\angle PSR$ . We then make use of the fact that the sum of the angles in a triangle is ${{180}^{\circ }}$ for triangle PRS to find the angle $\angle PRS$.

Complete step by step answer:
According to the problem, we are given that PQ is the diameter of the circle. If $\angle PQR={{65}^{\circ }}$ , $\angle RPS={{25}^{\circ }}$ and $\angle QPT={{60}^{\circ }}$ . We need to find the values of the angles (i) $\angle QPR$ , (ii) $\angle PRS$ , (iii) $\angle PSR$ and (iv) $\angle PQT$ .
Let us redraw the given figure,

From the figure, we can see that $\angle QTP$ is the angle formed by connecting the ends of the diameter of the circle, which means that the angle $\angle QTP$ is the angle in a semicircle. We know that angle in a semicircle is ${{90}^{\circ }}$ so, we get $\angle QTP={{90}^{\circ }}$ ---(1).
We know that the sum of the angles in a triangle is ${{180}^{\circ }}$.
From the triangle $PQT$ , we have $\angle QTP+\angle TPQ+\angle PQT={{180}^{\circ }}$ .
From equation (1), we get
$\Rightarrow {{60}^{\circ }}+{{90}^{\circ }}+\angle PQT={{180}^{\circ }}$ .
$\Rightarrow {{150}^{\circ }}+\angle PQT={{180}^{\circ }}$ .
$\Rightarrow \angle PQT={{30}^{\circ }}$ .
From the figure, we can see that $\angle QRP$ is the angle formed by connecting the ends of the diameter of the circle, which means that the angle $\angle QRP$ is the angle in a semicircle. We know that angle in a semicircle is ${{90}^{\circ }}$ , so we get $\angle QRP={{90}^{\circ }}$ ---(2).
We know that the sum of the angles in a triangle is ${{180}^{\circ }}$.
From the triangle $PQR$ , we have $\angle PQR+\angle QRP+\angle QPR={{180}^{\circ }}$ .
From equation (2), we get
$\Rightarrow {{65}^{\circ }}+{{90}^{\circ }}+\angle QPR={{180}^{\circ }}$ .
$\Rightarrow {{155}^{\circ }}+\angle QPR={{180}^{\circ }}$ .
$\Rightarrow \angle QPR={{25}^{\circ }}$ .
We can see that the points P, Q, R, S form a concyclic quadrilateral. We know that the sum of opposite angles in a concyclic polynomial is ${{180}^{\circ }}$.
So, we have $\angle PQR+\angle PSR={{180}^{\circ }}$ .
$\Rightarrow {{65}^{\circ }}+\angle PSR={{180}^{\circ }}$ .
$\Rightarrow \angle PSR={{115}^{\circ }}$ ---(3).
We know that the sum of the angles in a triangle is ${{180}^{\circ }}$.
From triangle $PSR$ , we have $\angle PSR+\angle PRS+\angle RPS={{180}^{\circ }}$ .
From equation (3), we get
$\Rightarrow {{25}^{\circ }}+{{115}^{\circ }}+\angle PRS={{180}^{\circ }}$ .
$\Rightarrow {{140}^{\circ }}+\angle PRS={{180}^{\circ }}$ .
$\Rightarrow \angle PRS={{40}^{\circ }}$ .
$\therefore$ We have found the values of angles $\angle QPR$ , $\angle PRS$ , $\angle PSR$ and $\angle PQT$ as ${{25}^{\circ }}$ , ${{40}^{\circ }}$ , ${{115}^{\circ }}$ and ${{30}^{\circ }}$ .

Note:
We can see that the given problem contains a huge amount of calculations, so we need to perform each step carefully to avoid confusion. We should not confuse angles while using the property of angle in a semi-circle. We should not make calculation mistakes while solving this problem. Whenever we get this type of problem, we should make use of properties of concyclic polygons and angle in a semi-circle to get the required answer.