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ashish_banga (1016)

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if PQ and RS be tangents at the extrimities of diameter PR of a circle of radius r . if PS and RQ intersect at a point X on the circumference of the circle, then 2r equals

    
spideyunlimited (4216)

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sqrt [ (PX)^2 + (RX)^2 ]

= sqrt [ (PQ)^2 + (RS)^2 - (QS)^2 ]

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- Gaurav Ragtah (spideyunlimited)
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ashish_banga (1016)

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no the answer is rt ( PQ . RS )
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budokai_tenkaichi_returns (409)

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solve dis


i kwufv


SAIYANS ARE OF TRUE WARRIOR RACE . DONT UNDER-ESTIMATE US!
special theory of relativity ....i luv it...
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seraphic (133)

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if PQ and RS be tangents at the extrimities of diameter PR of a circle of radius r . if PS and RQ intersect at a point X on the circumference of the circle, then 2r equals


in triangle PRS


tan(90-@) = RS/2r => cot@ = RS/2r  ...... eq. 1


in triangle PQR


tan@ = PQ/2r ...... eq. 2


from the above eq. 1 and 2


(2r)2 = RS.PQ


 


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