L. of Cosines?

Question

Hi.
I have this problem I keep doing wrong. It involves the law of cosines and sines.
A triangel with vertices A, B and C and opposite A we have side a, opposite B we have side b and opposite C we have side c. The length of b is 147, length of c is 216 and the angle A is 83,33°. I need to find B, C, a and the area of this triangle.

Could someone please walk me through the solution step by step (preferably with some explanations) so I can see where I made the mistake.
Thank you.

Answer 1

a^2 = b^2 + c^2 - 2bc cos(A)
a^2 = 147^2 + 216^2 - 2*147*216cos(83.33deg)
a^2 = (21609 + 46656) - 63504*0.1161
a^2 = 68265 - 7376.03
a^2 = 60888.97
a = sqrt(60888.97) = 246.76

sin(C) = 216 sin(83.33) / 246.76
= 0.8694
C = 60.39deg.

B = 180deg - (83.33 + 60.39) = 36.28deg.

Area = 147 * 216 * sin(83.33) / 2
= 31752 * 0.9932 / 2
= 15768.54.

Answer 2

According to the Law of Cosines, a^2 = b^2 + c^2 - 2bc cos A. Since b, c and A are given, this expression gives a. Now that you have a, b and C, you can use the L. of cosines again to find the cosines of B and C and then find B and C. But I think it's easier to use the L. of sines, that is

aA/sin(A) = b/sin(b) = c/sin(C) = D, D the diameter of the circumscribed circle (in this case, D doesn't matter).

And since you have a, b and C, you can compute the area S by the formula S = sqrt(p(p-a)(p-b)(p-c)), where p = (a+b+c)/2 is the triangle semi-perimeter.

Answer 3

The area is easiest, since you are given two adjacent sides (b,c) and the angle between them (A):

Area = (1/2)·b·c·sin(A)

Next, from the Law of Cosines:

a² = b² + c² – 2bc(cos(A))

That gives you "a" directly, since you already know b, c and angle A.

Once you know "a", you can use the Law of Sines to figure out angles B and C:

sin(A)/a = sin(B)/b = sin(C)/c

(i.e.: sin(B) = b·sin(A)/a; and sin(C) = c·sin(A)/a).

Answer 4

I don't think that your triangle is a right triangle, so you shouldn't be using sine and cosines.