anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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AravindG
  • AravindG
a fig would be helpful
anonymous
  • anonymous
But figure is not given.
anonymous
  • anonymous
Just try it out. I believe figure is not that difficult.

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anonymous
  • anonymous
Someone reply me plzzzzzzzzzzzzzz.
AravindG
  • AravindG
srry i gotta go help ya latr
anonymous
  • anonymous
option d is correct...
anonymous
  • anonymous
Okay. As ur wish. But others p[lzzzzzzzzzz try it out.
anonymous
  • anonymous
How?
anonymous
  • anonymous
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
anonymous
  • anonymous
O is the center of the circle
anonymous
  • anonymous
Thanks a lot
anonymous
  • anonymous
Brother, one physics doubt. Help please.

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