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Question

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
can't be determined
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B
65
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C
30
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D
75
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Solution

The correct option is B 65
ROQ = 180 - 50 = 130
( OQR+ORP+QOR+QPR=360and OQP=ORP=90)
Now, Since RT = TM and QS = SM
also OR = OM = OQ
ROT=TOM and MOS=SOQ SOT=12ROQ SOT=1302=65

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