Question

In the given figure O is the centre of the circle and $ \angle PAQ = {35^\circ } $ , then $ \angle OPQ $ is
$
  \left( {\text{A}} \right){75^\circ } \\
  \left( {\text{B}} \right){65^\circ } \\
  \left( {\text{C}} \right){85^\circ } \\
  \left( {\text{D}} \right){55^\circ } \\
  $

Answer 1

Hint: The diagram is the important part of the given problem. First try to use the theorem of the angle subtended by an arc of a circle to find the angle which is at the centre of the circle and then try to use the properties of isosceles triangle and other properties if required to find the angles that are inside the triangle.

Complete step-by-step answer:
The figure given to us is the main observation point.
As we can observe $ \angle PAQ $ and $ \angle POQ $ are subtended by the same arc.
Hence we know that “ Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the centre by the same arc”
Hence by the above theorem we can conclude that $ \angle POQ $ =2 $ \angle PAQ $
Since we know that $ \angle PAQ = {35^\circ } $
$ \Rightarrow \angle POQ = {70^\circ } $
now we have to find $ \angle OPQ $
Consider the triangle POQ.
Since OP and OQ are the radius of the circle hence they are equal.
Hence triangle POQ is an isosceles triangle. And with sides OP and OQ equal.
Hence according to the property of isosceles triangle, we know that $ \angle OPQ = \angle OQP - - - \left( 1 \right) $
Now we also know that the sum of angles of a triangle is $ {180^\circ } $ .
But $ \angle POQ = {70^\circ } $ .
Hence $ \angle OPQ + \angle OQP = 180 - 70 = 110 $
Using (1) we get that $ \angle OPQ = \angle OQP = {55^\circ } $
Hence we get that $ \angle OPQ = {55^\circ } $
hence option (D) is the right answer.
So, the correct answer is “Option D”.

Note: In this type of sums we usually have to use different theorems of the circle so that we can approach our aim correctly. Apart from the circle theorem and properties, one has to use different geometrical properties as used in the sum above to obtain the desired result and values.