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

15^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60^{o}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

45^{o}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $45^{o}$
$P R = Q R$ ....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.

So, option D is correct.

Suggest Corrections

Similar questions

{cos}^{2} 75^{o} + {cos}^{2} 45^{o} + {cos}^{2} 15^{o} - {cos}^{2} 30^{o} - {cos}^{2} 60^{o}

\frac{x {cosec}^{2} 30^{o} {sec}^{2} 45^{o}}{8 {cos}^{2} 45^{o} {sin}^{2} 90^{o}} = {tan}^{2} 60^{o} - {tan}^{2} 45^{o}

x

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

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

\frac{2}{3} ({cos}^{4} 30^{o} - {sin}^{4} 45^{o}) - 3 ({sin}^{2} 60^{o} - {sec}^{2} 45^{o}) + \frac{1}{4} {cot}^{2} 30^{o}

Join BYJU'S Learning Program