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

Answer 1

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

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!

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