Triangle question?

Question

This question seems very easy, I just don't know how to do it. please help me out...

Solve the triangle with the given dimensions ( angle measures): a=5, b=7, c=6

**the given numbers are side measures, I must find angle measures.

thanks everyone, i got...
A= 44.415
B= 78.463
C= 57.122

Answer 1

Use the law of cosines:

a^2 = b^2 + c^2 - 2bc*cosA
b^2 = a^2 + c^2 - 2ac*cosB
c^2 = a^2 + b^2 - 2ab*cosC

cosA = (a^2 - b^2 - c^2) / (-2bc)
A = arccos[(a^2 - b^2 - c^2) / (-2bc)]
A = arccos[(25 - 49 - 36) / (-2*7*6)]
A = arccos(.714285714286)
A = 44.4153 degrees

cosB = (b^2 - a^2 - c^2) / (-2ac)
B = arccos[(b^2 - a^2 - c^2) / (-2ac)]
B = arccos[(49 - 25 - 36) / (-2*5*6)]
B = arccos(.2)
B = 78.463 degrees

C = 180deg - A - B
C = 180 - 44.4153 - 78.463
C = 57.1217 degrees

Answer 2

use the law of cosines to find one of the angles, then the law of sines for one of the remaining angles, then subtract the sum from 180 for the remaining angle.

Answer 3

a^2 = b^2 +c^2 - 2bc cosA
2bc cosA = b^+c^-a^2
cosA= (b^2+c^2 -a^2)/(2bc)
A = arccos[(b^2+c^2 -a^2)/(2bc)]

Similarly,
B= arccos[(a^2+c^2-b^2)/(2ac)]

Then you can find C by C = 180 - (A+B)

Answer 4

A= 44°24'55"
B= 78°27'47"
C= 57°07'18"

oops.... this one's right.