PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that $∠$ POR $= 120^{\circ}$ , then $∠$ OPQ is $30^{\circ}$ .

60°

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45°

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Solution

The correct option is A
60°

Given that $∠ P O R = 120^{\circ}$

We know that $∠$ OQP is $90^{\circ}$

Using external angle theorem, $∠ P O R = ∠ O Q P + ∠ O P Q$

$\Rightarrow 120^{\circ} = 90^{\circ} + ∠ O P Q$

$\Rightarrow ∠ O P Q = 120^{\circ} - 90^{\circ}$

$\Rightarrow ∠ O P Q = 30^{\circ}$

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PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that ∠POR =120∘, then ∠OPQ is 30∘.

The correct option is A 60° Given that ∠POR=120∘ We know that ∠OQP is 90∘ Using external angle theorem, ∠POR=∠OQP+∠OPQ ⇒120∘=90∘+∠OPQ ⇒∠OPQ=120∘–90∘ ⇒∠OPQ=30∘

PQ is a tangent drawn from a point P to a circle with center O and QOR is a diameter of the circle such that $∠$ POR $= 120^{\circ}$ , then $∠$ OPQ is $30^{\circ}$ .