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In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If `/_QPT=60^@` find `/_PRQ` |
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Answer» In the given figure, `/_QPT = 60^@` `/_QPO = /_OPT-/_QPT = 90-60 = 30^@` As, `OP = OQ`, `:. /_OPQ =/_OQP = 30^@` `:. /_POQ = 180-30-30 = 120^@` Now, `/_PRQ = 1/2(360-/_POQ) = 240/2 = 120^@` |
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