## anonymous one year ago In right ∆ABC, angle B = 90°, angle C = 40°, and angle BC = 10. What are the other two side lengths of the triangle? A) AC = 13.1, AB = 8.4 B) AC = 18.7, AB = 15.8 C) AC = 14.3, AB = 10.2 D) AC = 15.6, AB = 11.9

1. anonymous

Could really use the help, don't have to give me the answer outright just need to know how to solve this problem, can someone please guide me in the right direction? thank you.

2. Nnesha

|dw:1434457799463:dw| $\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

3. Nnesha

so you need one angle and one side to find the other side of right triangle after that you can apply Pythagorean theorem

4. anonymous

@Nnesha Ahh so I solve it using SOHCAHTOA which is what my teacher taught me awhile back, I haven't done this sort of stuff in forever, might need some more assistance haha

5. Nnesha

yes alright!

6. Nnesha

as you can there is one angle and 2 sides in SOHCAHTOA formula i posted above so you need one angle and one side to solve for 3rd variable cos = adjacent over hyp so we to use cos function $\huge\rm \cos (40) = \frac{ 10 }{ h }$ h = hypotenuse solve for h

7. Nnesha

|dw:1434458317996:dw| adjacent side that touch the angle

8. anonymous

@Nnesha I got D as my answer

9. Nnesha

how did you get D ?

10. Nnesha

$\huge\rm \cos (40) = \frac{ 10 }{ h }$ solve for h and it's 40 degree

11. Nnesha

do you know how to solve for h ??

12. anonymous

@Nnesha Yea, I ended up getting 7.66

13. Nnesha

what is cos(40) make sure mode is 'n degree

14. anonymous

@Nnesha I got cos(40)=0.76604=10/h then I rearranged that, then multiplied 0.7660 x 10 and that was what I got, im trying to get an idea from this

15. Nnesha

why did multiply ??

16. Nnesha

you*

17. anonymous
18. Nnesha

what about it ?

19. Nnesha

$\huge\rm .766=\frac{ 10 }{ h }$ you should multiply both sides by H not by 10

20. Nnesha

$\huge\rm .766 \times h = \frac{ 10 }{\cancel h} \times \cancel{h}$

21. anonymous

@Nnesha Then divide both sides by .766? I ended up getting 13.054 or 13.1

22. Nnesha

yes right

23. Nnesha

right |dw:1434460279833:dw| now you can apply Pythagorean theorem to find 3rd side of right triangle $\huge\rm a^2 +b^2 =c^2$ plug in values and solve for 3rd side remember c is the longest side of right triangle which is hypotenuse

24. anonymous

alright I did 10^2+b^2=13.1^2 then got 100+b^2=171.61 then subtracted 100 from both sides got b^2=71.61, I squared both sides and got B=8.4622

25. Nnesha

yes right

26. Nnesha

b^2 = 71.61 so b=what ?

27. anonymous

8.4622

28. Nnesha

right

29. anonymous

so its A, thank you so much haha, idk if you would be willing to help me with one more problem considering how long it took awhile to complete this haha

30. Nnesha

i'll try my best :-)

31. anonymous

@Nnesha haha ok thank you, I appreciate it In ∆ABC, angle B is a right angle and tan A = 13/16 . What is angle C ?

32. Nnesha

what's the definition of tan ?? look at my first comment tan equal what ?

33. anonymous

34. Nnesha

yes now draw a right triangle

35. anonymous

@Nnesha Alright done

36. Nnesha

|dw:1434461030870:dw| which side is 13 and which one is 16 ?

37. anonymous

opposite should be 13 and adjacent should be 16?

38. Nnesha

yes right u should write 13 and 16 ?

39. anonymous

@Nnesha It looks funny to me, the adjacent side is longer than the opposite side?

40. Nnesha

that doesn't matter hypotenuse should be bigger than other 2 sides doesn't matter a is longer or b

41. Nnesha

which side is opposite of angle A ?

42. anonymous

@Nnesha the side that's 13

43. Nnesha

yes so |dw:1434461354773:dw|where u should write 13 ? and where suppose to be 16 ?

44. anonymous

between A and B would be the 13 and between B and C would be the 16

45. Nnesha

|dw:1434461430223:dw| yep right now tanC = opposite over adjacent substitute values solve for C

46. anonymous

@Nnesha Im having difficulty understanding what to do >_< would I multiply both sides by one of the numbers or?

47. Nnesha

did u substitute values of opposite and adjacent side let me know u get after that

48. anonymous

Ok yea I just push in Tan(C)=13/16

49. Nnesha

well wait we did it wrong

50. Nnesha

tan A = 13/16 not tan C

51. Nnesha

so it should be like this |dw:1434462085204:dw|

52. anonymous

@Nnesha Ahhh ok, I knew something was off, so then Tan(C)=16/13

53. Nnesha

now tan = opposite over adjacent $\huge\rm tan (C)= \frac{ 16 }{ 13 }$ yes that's right

54. Nnesha

so u need to find invs of tan $\huge\rm C = \tan^{-1} \frac{ 16 }{ 13 }$ use calculator

55. anonymous

Alright, I got 0.8884

56. Nnesha

make sure mode 'n degree

57. anonymous

@Nnesha Yea, it keeps giving me the same answer for some reason C=0.88847979

58. Nnesha

bec mode of calculator is on rad so it should be on degree

59. Nnesha

hit rad and then type that

60. Nnesha
61. anonymous

I got this now C=50.906141113771

62. anonymous

@Nnesha

63. Nnesha

looks good

64. anonymous

@Nnesha Thank you for everything, I really appreciate it

65. Nnesha

my pleasure