## BloomLocke367 one year ago Find the area.

1. anonymous

what

2. anonymous

whats the question

3. BloomLocke367

|dw:1433197250937:dw|

4. BloomLocke367

@Nnesha

5. anonymous

hold on

6. BloomLocke367

@perl

7. anonymous

okay this link will help https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

8. BloomLocke367

@Nnesha

9. Nnesha

http://prntscr.com/7c1xmi you need to find 3rd side of triangle it's not a right triangle so law of cosine ?familiar with this method ?

10. BloomLocke367

kind of, I sort of know what it is, but I don't really know how to use it

11. BloomLocke367

I have to eat dinner, be back in a sec

12. Nnesha

alright lol

13. Nnesha

have fun!

14. BloomLocke367

I'm back, I wasn't that hungry

15. BloomLocke367

@Nnesha

16. Nnesha

ahh alright so did you eat candy ? :P

17. BloomLocke367

no, I had like a few bites.. I think it's my nerves that have me not hungry.. I'm very stressed.

18. Nnesha

$\huge\rm c^2 = a^2 +b^2 -2abcos C$ |dw:1433199653612:dw| plug in value solve for c

19. BloomLocke367

wait, don't I need to use law of sines for this?

20. Nnesha

mhm nope bec given is two sides and one angle

21. BloomLocke367

yes, you use cosine if you have 3 sides, and law of sines if you have 2 sides and an angle.. I'm almost positive

22. Nnesha

http://mathforum.org/library/drmath/view/69918.html you can use law of cosines to find 3rd side

23. BloomLocke367
24. Nnesha
25. BloomLocke367

oh, okay.. sorry for doubting you

26. Nnesha

nah it's fine! i like it :P

27. Nnesha

okay solve for c

28. BloomLocke367

c^2=64+324-2(8)(18)cos(25), so far, right?

29. BloomLocke367

c^2=388-288cos(25) c^2=100cos(25)

30. BloomLocke367

right?

31. Nnesha

hold on >:P

32. BloomLocke367

okay

33. Nnesha

yep right solve for c

34. BloomLocke367

I got c^2=90.63 so c=9.52

35. BloomLocke367

@Nnesha

36. BloomLocke367

@Nnesha

37. Nnesha

hmm i got 11.269

38. BloomLocke367

how? (sorry, I had to go to bed) @Nnesha

39. BloomLocke367

@pooja195

40. geerky42

Why do we need to find value of c? |dw:1433254516249:dw| Obviously, $$\large A_{\text{triangle}} = \dfrac12bh$$ We already have base $$\overline{\text{QR}} = 8$$ So we need to find h: Using fact that $$\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}}$$, we have $$\sin Q = \dfrac{h}{\overline{\text{PQ}}} = \dfrac{h}{18}$$ Yield $$18\sin(25^\text o) = h$$ So we have $$A_{\text{triangle}} = \dfrac12~\overline{\text{QR}}~h = \dfrac{1}{2}(8)(18\sin(25^\text o)~) = \cdots~?$$

41. BloomLocke367

I got 30.43

42. geerky42

Me too. So that's your area.

43. geerky42

Do you understand what I wrote?

44. BloomLocke367

oh, okay. kinda, I don't understand how you got the height, but the rest I get.

45. geerky42

well, we can set up height of triangle here:

46. geerky42

sorry hold on...

47. BloomLocke367

okay. Can you help with more?

48. geerky42

|dw:1433254921641:dw| Then we can treat it as "new triangle" |dw:1433254928278:dw| From here, we have hypotenuse and angle Q and height. So using sine, we can solve for h. Then we use that height to solve for area of original triangle.

49. geerky42

I can try.

50. BloomLocke367

ohhhhhhh okay. and it's the same height for both triangles?

51. geerky42

Same angle and hypotenuse, so yeah same height,

52. BloomLocke367

alright. I have some more questions, if that's okay. I'll post a new question and tag you

53. geerky42

sure