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Question

The given figure shown a circle with centre O. Also, PQ = QR = RS and PTS = 75o. Calculate :

(i) POS,

(ii) QOR,

(iii) PQR.

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Solution

Given PQ = QR = RS c8.jpg

\Rightarrow \angle POQ = \angle QOR = \angle ROS (Equal chords subtend equal angles at the center of a circle)

\angle POS = 2 \ \angle PTS (angle subtended at the center is twice that of the one subtended on the circumference)

\angle POQ = \angle QOR = \angle ROS = \frac{150}{3} = 50^o

In \triangle OPQ ,

OP = OQ (radius of the same circle)

\angle OPQ = \angle OQP

Therefore \angle OPQ + \angle OQP + \angle POQ = 180^o

\angle OPQ + \angle OQP = 180-50=130^o

2 \angle OPQ = 130

\angle OPQ = 65^o

Similarly, in \triangle OQR, \angle OQR = \angle ORQ = 65^o

and in \triangle ORS, \angle ORS = \angle OSR = 65^o

Therefore

(i) \angle POS = 150^o

(ii) \angle QOR = 50^o

(iii) \angle PQR = \angle PQO + \angle OQR = 65+65 = 130^o


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