PQ is a diameter of a circle whose centre is O. If a point R lies on the circle and $$\angle RPO$$ is $$39^o$$, then find the measure of $$\angle RQP$$.​

Correct option is A

Given:

$PQ$ is the diameter of the circle, with center $O$.

A point $R$ lies on the circle.

The angle ∠RPO=39∘\angle RPO = 39^\circ∠RPO  = 39∘39^\circ39∘​, and we need to find $\angle RQP$.  

Concept Used:

In a circle:

The angle subtended by a diameter at the circumference is$$90^\circ$$​ (right angle).

The sum of angles in △PRQ is $$180^\circ.$$​​

​Solution: 

​Since PQPQPQ is the diameter of the circle, and R is the point on the circle, So:

$$\angle PRQ = 90^\circ$$ 

$\angle RPO=$​​$$39^\circ$$.  

Now, 

$$\angle PRQ + \angle RPO + \angle RQP = 180^\circ \\\implies 90^\circ + 39^\circ + \angle RQP = 180^\circ \\\implies\angle RQP = 180^\circ - 90^\circ - 39^\circ \\\implies\angle RQP = 51^\circ$$​​​​