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If tangents QR, PR, and PQ are drawn respectively at A, B and C to the circle circumscribing an acute angled triangle triangle ABC so as to form another triangle PQR, then find the alternative equivalent to angle RPQ. a) Angle BAC b) 0.5( 180 degrees - angle BAC) c) 180 degrees - angle BAC d) (180 degrees - 2 times angle BAC)

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a fig would be helpful
But figure is not given.
Just try it out. I believe figure is not that difficult.

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Someone reply me plzzzzzzzzzzzzzz.
srry i gotta go help ya latr
option d is correct...
Okay. As ur wish. But others p[lzzzzzzzzzz try it out.
I am assuming that vertex P is opposite of A, i.e. it is created by tangents to B and C .. ( normal convention) .. Now when you complete the triangle PQR , concentrate on quadilateral OBCP ... here angle OBP = OCP = 90 and hence angle BOC = 180 - P. Now segment BC for the circle also suspends angle A on the circumference which would be half of what the segment suspends on the center (property of circle) .. hence (180-p)/2 = A ... hence p = 180 - 2A
O is the center of the circle
Thanks a lot
Brother, one physics doubt. Help please.

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