How do you find angles A, B, and C if in triangle ABC, #a=12#, #b=15#, and #c=20#?

1 Answer
May 3, 2018

Answer:

#A ~~ 36.71^@#
#B ~~ 48.34^@#
#C ~~ 94.93^@#

Explanation:

To solve the for the angles when we have the lengths of all three sides, we use the law of cosines.

The law of cosines states that angle #A = cos^-1((a^2 - b^2 - c^2)/(-2bc))#, #B = cos^-1((b^2 - a^2 - c^2)/(-2ac))#, and #C = cos^-1((c^2 - a^2 - b^2)/(-2ab))#.

Let's find angle #A# first:
#A = (a^2 - b^2 - c^2)/(-2bc)#

#A = (12^2 - 15^2 - 20^2)/(-2(15)(20))#

#A = cos^-1((-481)/(-600))#

#A ~~ 36.71^@#

Now angle #B#:
#B = cos^-1((b^2 - a^2 - c^2)/(-2ac))#

#B = cos^-1((15^2 - 12^2 - 20^2)/(-2(12)(20)))#

#B = cos^-1((-319)/(-480))#

#B ~~ 48.34^@#

Finally angle #C#:
#C = cos^-1((c^2 - a^2 - b^2)/(-2ab))#

#C = cos^-1((20^2 - 12^2 - 15^2)/(-2(12)(15)))#

#C = cos^-1((31)/(-360))#

#C ~~ 94.93^@#

We can also find angle #C# by doing #180^@ - 36.71^@ - 48.34^@#, since the measures of the angles in a triangle add up to #180^@#.

Hope this helps!