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Aadarsh

  • 2 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)

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  1. AravindG
    • 2 years ago
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    a fig would be helpful

  2. Aadarsh
    • 2 years ago
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    But figure is not given.

  3. Aadarsh
    • 2 years ago
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    Just try it out. I believe figure is not that difficult.

  4. Aadarsh
    • 2 years ago
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    Someone reply me plzzzzzzzzzzzzzz.

  5. AravindG
    • 2 years ago
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    srry i gotta go help ya latr

  6. shaan_iitk
    • 2 years ago
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    option d is correct...

  7. Aadarsh
    • 2 years ago
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    Okay. As ur wish. But others p[lzzzzzzzzzz try it out.

  8. Aadarsh
    • 2 years ago
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    How?

  9. shaan_iitk
    • 2 years ago
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    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
    • 2 years ago
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    O is the center of the circle

  11. Aadarsh
    • 2 years ago
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    Thanks a lot

  12. Aadarsh
    • 2 years ago
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    Brother, one physics doubt. Help please.

  13. Aadarsh
    • 2 years ago
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    http://openstudy.com/study#/updates/4f25694ce4b0a2a9c266e802

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