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^@#