# Question #a69e0

##### 1 Answer

Nov 18, 2016

#### Explanation:

By convention,

The law of cosines states that

#b^2 = a^2+c^2-2ac cos(B)# #a^2 = b^2+c^2-2bc cos(A)#

If we try solving for the angle, we get

#cos(A) = (b^2+c^2-a^2)/(2bc)# #cos(B) = (a^2+c^2-b^2)/(2ac)# #cos(C) = (a^2+b^2-c^2)/(2ab)#

Thus, applying the inverse cosine function to both sides:

#A = arccos((b^2+c^2-a^2)/(2bc))# #B = arccos((a^2+c^2-b^2)/(2ac))# #C = arccos((a^2+b^2-c^2)/(2ab))#

Substituting in the given values

#A~~44.2^@# #B ~~ 95.5^@# #C ~~ 40.3^@#