Question

PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a dimeter . if angle POR is 130 and S is point on the circle , find angleQPO + angle RST

Answer 1

HERES THE SOLUTION OF YOUR QUESTION.

Answer 2

Answer:

The sum of ∠QPO + ∠RST is 105°

Step-by-step explanation:

Given PQ is a tangent from an external point P to a circle with center O and OP cuts the circle at T and QOR is a diameter. If ∠POR is 130° and S is point on the circle. we have to find the sum of ∠QPO + ∠RST

In ΔPOQ, by exterior angle property of triangle

∠ROT=∠OQP+∠OPQ

 130° = 90°+ ∠OPQ    (∵ the point of contact of tangent and radius makes 90°)

∠OPQ=130°-90°=40°

By the theorem which states that angle subtended at the center is twice the angle at the circumference of circle.

$\angle ROT=2\angle RST$

⇒ $\angle RST=\frac{1}{2}\angle ROT=\frac{1}{2}\times 130^{\circ}=65^{\circ}$

Hence, ∠QPO + ∠RST=40°+65°=105°