## anonymous one year ago Will Medal! The measure of theta, in degrees, is approximately 48.8 degrees? A. True B. False (Picture Below) ***My Answer: B***

1. anonymous

2. UsukiDoll

do have any reasons why you think it's B?

3. UsukiDoll

did you use the law of Cosines? cos A = $\cos A = \frac{(b^2+c^2-a^2)}{2bc}$ $\cos B = \frac{(c^2+a^2-b^2)}{2ca}$

4. anonymous

I used the formula: $c ^{2} = a ^{2} + b ^{2} -2ab cosC$

5. zeesbrat3

Couldn't you just use arccos?

6. UsukiDoll

hmmm but I want cos C to be by itself and it's easier to grab the angle A and B first.

7. anonymous

Now that I see the formulas you gave me, I think I may be doing the wrong formula.

8. UsukiDoll

@zeesbrat3 arccos is later into the formula... you have to find angle A and angle B first .. since this is all S S S (side side side)

9. zeesbrat3

Fair enough

10. UsukiDoll

c is supposed to be the longest side a.k.a hypotenuse

11. anonymous

Oh ok. The laws of cosines were getting me confused for a second.

12. anonymous

So we would have a = 5.3 b = 4 and c = 7?

13. UsukiDoll

I need letters first... we need to redraw the triangle with A B C and then look across for our a b c |dw:1434936158533:dw| ok let's use those values

14. UsukiDoll

so that would mean ... a = 5.3 , b=4, and c = 7... now we use Cos A to find Angle A

15. anonymous

Ok

16. UsukiDoll

$\cos A = \frac{(b^2+c^2-a^2)}{2bc}$ you just have to plug in those values.. and simplify

17. anonymous

Great! Thank you! So we would have cos A = 0.66?

18. anonymous

Then in degrees it would be 37.8 degrees?

19. UsukiDoll

this is where you need to use $\cos^{-1}(0.66)$ and then you use a calculator to find the degrees of it

20. UsukiDoll

Can I just double check this for a sec? ^_^ so fast

21. anonymous

Oh ok! Yeah go ahead :)

22. UsukiDoll

$\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}$

23. anonymous

Oh nevermind I got 48.7 degrees

24. zeesbrat3

I got $\cos \theta = .6835$

25. UsukiDoll

$\cos A = \frac{(16+49-28.09)}{56} \rightarrow \frac{36.91}{56} \rightarrow 0.659 \rightarrow 0.66$

26. UsukiDoll

$\cos^{-1}(0.66) \rightarrow 48.7$ I also got 48.7 degrees

27. zeesbrat3

My bad, I used 54 as the denominator, not 56... ignore the blonde...

28. UsukiDoll

ok cool... we got angle A = 48.7 degrees... Now we need angle B

29. anonymous

Oh whoops for the the rest of the equation haha

30. UsukiDoll

let's try and solve further.

31. anonymous

Ok

32. UsukiDoll

$\cos B = \frac{(c^2+a^2-b^2)}{2ca}$ c = 7, b = 4, a = 5.3

33. UsukiDoll

back the equation tool gave me a hard time

34. anonymous

I got around 35.1 degrees (34.95 not rounded)

35. UsukiDoll

$\cos B = \frac{(7^2+5.3^2-4^2)}{2(7)(5.3)}$ $\cos B = \frac{(49+28.09-16)}{74.2}$ $\cos B = \frac{61.09}{74.2}$ $\cos B = 0.84$ $\cos^{-1}(0.84) =31.95 -> 32$ huh that's odd... how did you get 35.1 degrees?

36. triciaal

another approach when given only the 3 sides use cosine rule to find largest angle opposite longest side

37. UsukiDoll

oh wait I got the division wrong $\frac{61.09}{74.2} = 0.823$

38. UsukiDoll

$\cos^{-1}(0.823) = 34.58$

39. UsukiDoll

anyway .. let's just assume that Angle B = 34.58 Angle A = 48.7 to find Angle C.. we use the fact that angles of a triangle must add up to 180 degrees Angle C = 180-48.7-34.58 I will also take the 35.1 into consideration for Angle B Angle C = 180-48.7-35.1

40. triciaal

|dw:1434937131238:dw|

41. triciaal

now use the sin rule

42. UsukiDoll

Assuming that Angle B is 34.58 Angle C = 180-48.7-34.58 = 96.72 Assuming that Angle B is 35.1 Angle C = 180-48.7-35.1=96.20 But I think ... in the drawing... only one angle is being considered.

43. triciaal

|dw:1434937491614:dw|

44. triciaal

With rounding error true

45. UsukiDoll

I had angle A almost close to 48.8 as well because when I divided $\frac{36.91}{56}$

46. UsukiDoll

48.76 since the number after the decimal point is greater than 5 I can round up and it does become 48.8

47. UsukiDoll

if rounding isn't required then it wouldn't be 48.8 but that's cutting it close..