## Aadarsh 3 years ago 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)

1. AravindG

But figure is not given.

Just try it out. I believe figure is not that difficult.

5. AravindG

srry i gotta go help ya latr

6. shaan_iitk

option d is correct...

Okay. As ur wish. But others p[lzzzzzzzzzz try it out.

How?

9. shaan_iitk

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

10. shaan_iitk

O is the center of the circle

Thanks a lot