PR and PQ are tangents. $∠ Q O R$ is equal to

140^{\circ}

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120^{\circ}

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160^{\circ}

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170^{\circ}

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Solution

The correct option is B $120^{\circ}$
$∠ Q O R = ˚ 120$ (As per the theoram,angle at centre is double the angle at tangent.
since angle P is 60 degree angle o will be $120^{0}$

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\begin{matrix} H e i g h t (i n c m) & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 & 150 - 160 & 160 - 170 & 170 - 180 N o . o f g o a l s s c o r e d & 8 & 10 & 15 & 9 & 12 & 14 & 13 & 10 \end{matrix}

P Q

P R

O

∠ P = \frac{4}{5} ∠ O .

∠ Q O R

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) $∠ Q S R,$

154

Classes	$120 - 130$	$130 - 140$	$140 - 150$	$150 - 160$	$160 - 170$	$170 - 180$
Frequency	$2$	$8$	$12$	$f$	$8$	$7$

∠ Q P R = 46^{\circ}

∠ Q O R

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