In the adjoining figure, 'O' is the centre of the circle and PQ , PR and ST are the three tangents. ∠QPR=50∘, then the value of ∠SOT is
A
30∘
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B
75∘
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C
65∘
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D
can't be determined
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Solution
The correct option is C65∘ ∠ROQ = 180 - 50 = 130∘ (∵∠OQR+∠ORP+∠QOR+∠QPR=360∘and∠OQP=∠ORP=90∘) Now, Since RT = TM and QS = SM ∴ also OR = OM = OQ ∴∠ROT=∠TOMand∠MOS=∠SOQ∴∠SOT=12∠ROQ∴∠SOT=1302=65∘