Law of Cosine?

Question

Solve the triangle: a = 5, b = 8, c = 12

I found three angles, but their sum come out to 94.11 when it should be 180.

Where am I doing it wrong? I used the law of cosine.

Thanks.

I know what the formula is. As I said, I already tried this, but came out with a sum of angles equality 94 when the sum should be 180.

What are the correct angles?

Answer 1

c^2 = a^2 + b^2 -2ab cos C
144= 25 + 64 - 80 cos C
144- 64-25= -80 cos C
55/(-80) = cos C
cos C = -11/16

a^2 = b^2 + c^2 - 2bc cos A
25= 64 + 144 - 192 cos A
25-64-144 = -192 cos A
-183 = -192 cos A
cos A = (-183)/(-192)
cos A= 61/64

b^2 = a^2 +c^2 - 2ac cos B
64 = 25 + 144 - 120 cos B
64-25-144 = -120 cos B
-105 = -60 cos B
cos B = (-105)/(-120)
cos B = 21/24

if you look at the cosines you can see that one of the cosines is very close to cosine of π which is -1 (C) and the other 2 are very close to cosine of 0 which is 1 (A and B). This makes sense because if one of the angles is close to 180 then the other 2 should be very small angles.

Answer 2

Hah I learned this today!


Uhm, It's C^2=a^2+b^2-2(a)(b)cosC

So it would be..oh wait angles.....uhm....

CosC=a^2+b^2-c^2/2ab

Cos C= 5^2+8^2-12^2 / 2(5)(8)

Cos C= 25 +64 - 144 / 80

Cos C= .6875

Cos C is now multpylied by Cos ^-1 and so is the other side...

angle C = 133.4325366
---------------------------------

Cos B = 25 + 144 - 64 / 120

Cos B = .875

Cos^-1

Angle B = 28.95502437
-----------------

So add the angles and sutract from the triangle....

B+C= 162.387561 - 180

Angle A is 17.61243903
-------------------------------
And there are your angles.

A=17.61243903
B=28.95502437
C=133.4325366

Answer 3

angle c = inverse cos[(a squared + b squared minus c squared) / 2ab]

Answer 4

c^2 = a^2 + b^2 - 2ab·cos(µ)