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parvathy

  • 3 years ago

a quadrilateral ABCD,<B=130,<c=60,angle bisectors of<Aand<D meet at P.FIND<APD.

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  1. parvathy
    • 3 years ago
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    |dw:1329669343302:dw|

  2. parvathy
    • 3 years ago
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    answer is 95

  3. parvathy
    • 3 years ago
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    wht r u doing?

  4. parvathy
    • 3 years ago
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    answer is 95

  5. parvathy
    • 3 years ago
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    but answer is 95 knw

  6. parvathy
    • 3 years ago
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    ????????????????????????????????????????????????????????????????????????

  7. parvathy
    • 3 years ago
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    k

  8. parvathy
    • 3 years ago
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    k

  9. 2bornot2b
    • 3 years ago
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    |dw:1329670418181:dw| We get two equations from the figure \[x+y+p=180--------(1)\]\[x+y+(360-p)+130+60=540\implies x+y-p=-10-------(2)\]

  10. 2bornot2b
    • 3 years ago
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    \[2p=\frac{190}{2}------------[(1)-(2)]\]\[p=95\]

  11. parvathy
    • 3 years ago
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    we will plz mark the angles which i hav given in ma figure

  12. 2bornot2b
    • 3 years ago
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    |dw:1329670804174:dw| \[\huge labelled\]

  13. parvathy
    • 3 years ago
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    how u got 2p

  14. parvathy
    • 3 years ago
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    190?

  15. parvathy
    • 3 years ago
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    got

  16. 2bornot2b
    • 3 years ago
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    I subtracted the two equations. Note, first I created two equations, and then I subtracted the second from the first.

  17. parvathy
    • 3 years ago
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    2P?

  18. parvathy
    • 3 years ago
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    K

  19. parvathy
    • 3 years ago
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    U R VERY BRILLIANT

  20. 2bornot2b
    • 3 years ago
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    |dw:1329671057692:dw|

  21. parvathy
    • 3 years ago
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    k brilliant boy

  22. 2bornot2b
    • 3 years ago
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    Did you understand that?

  23. parvathy
    • 3 years ago
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    ya thnkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

  24. parvathy
    • 3 years ago
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    another question is coming

  25. 2bornot2b
    • 3 years ago
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    OK, just post it as a new question.

  26. parvathy
    • 3 years ago
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    k

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spraguer (Moderator)
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