If $∠ R P Q = 45^{o}$ , find $∠ P R Q$ .

60^{o}

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90^{o}

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120^{o}

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150^{o}

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Solution

The correct option is B $90^{o}$
$P R = Q R$ ....Tangents from an external point to the circle are equal.
$∴ △ R P Q$ is an isosceles triangle.
$∴ ∠ R P Q = ∠ P Q R = 45^{o}$ ...base angles of isosceles triangle.
$∠ P R Q + ∠ R Q P + ∠ Q P R = 180^{o}$ ...angle sum property of triangle.
$∠ P R Q = 180 - 45^{o} - 45^{o} = 90^{o}$ .
So, option B is correct.

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4 ({sin}^{4} 30^{o} + {cos}^{4} 60^{o}) - 3 ({cos}^{4} 45^{o} - {sin}^{2} 90^{o})

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∠ P O Q = 120^{o}

∠ P R Q

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