In the figure, $P Q$ and $P R$ are tangents to circle with centre $A$ . If $∠ Q P A = 27^{o}$ , then find $∠ Q A R =$ .

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Solution

$\begin{matrix} Δ A Q P + Δ A R P A P = A P c o m m o n ∠ Q = ∠ R [90^{\circ}] A Q = A R [r a d i i] S o 1 ∠ Q P A = ∠ R P A = 27^{\circ} ∠ P + ∠ Q + ∠ A + ∠ R = 360^{\circ} 54 + 90 + ∠ A + 90 = 360^{\circ} ∠ A = 360^{\circ} - 180 - 54 ∠ A = 180 - 54 ∠ A = 126^{\circ} ∠ Q A R = 126^{\circ} A n s . \end{matrix}$

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In the figure PQ and PR are tangents to the circle with centre ‘O’. If , shown that PQOR is a square.

In the following figure, PQ and PR are tangents to the circle, with centre O. If $∠ Q P R = 60^{o}$ , calculate :

(i) $∠ Q O R,$
(ii) $∠ O Q R,$
(iii) $∠ Q S R,$

∠ Q P R = 60^{\circ}

∠ Q R O = 30^{\circ} .

O

∠ P = \frac{4}{5} ∠ O .

∠ Q P R

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