Question

PQ is tangent to the circle with centre O such that OP = 20Q. m∠OPQ is

विकल्प

• 15°

• 60°

• 45°

• 30°

Answer 1

30°

Explanation:

1. Identify triangle properties: In the circle with centre O, OQ is the radius and PQ is the tangent at point Q. By circle geometry theorems, the radius is perpendicular to the tangent at the point of contact, making ∠OQP = 90°.

2. Use trigonometric ratios: Triangle OQP is a right-angled triangle. We are given the relationship OP = 2OQ. In this triangle, OQ is the side opposite to ∠OPQ and OP is the hypotenuse.

3. Calculate the Angle:

`sin(∠OPQ) = "Opposite"/"Hypotenuse"`

= `(OQ)/(OP)`

Substituting the given relationship OP = 2OQ:

`sin(∠OPQ) = (OQ)/(2OQ)`

= `1/2`

We know that `sin (30^circ) = 1/2`.

Therefore, ∠OPQ = 30°.