## parvathy 3 years ago a quadrilateral ABCD,<B=130,<c=60,angle bisectors of<Aand<D meet at P.FIND<APD.

1. parvathy

|dw:1329669343302:dw|

2. parvathy

3. parvathy

wht r u doing?

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5. parvathy

6. parvathy

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7. parvathy

k

8. parvathy

k

9. 2bornot2b

|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

$2p=\frac{190}{2}------------[(1)-(2)]$$p=95$

11. parvathy

we will plz mark the angles which i hav given in ma figure

12. 2bornot2b

|dw:1329670804174:dw| $\huge labelled$

13. parvathy

how u got 2p

14. parvathy

190?

15. parvathy

got

16. 2bornot2b

I subtracted the two equations. Note, first I created two equations, and then I subtracted the second from the first.

17. parvathy

2P?

18. parvathy

K

19. parvathy

U R VERY BRILLIANT

20. 2bornot2b

|dw:1329671057692:dw|

21. parvathy

k brilliant boy

22. 2bornot2b

Did you understand that?

23. parvathy

ya thnkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

24. parvathy

another question is coming

25. 2bornot2b

OK, just post it as a new question.

26. parvathy

k