Solve the triangle? when A = 24.3 B = 14.7 C = 18.7

1 Answer
Apr 26, 2018

Vertices:

#A = arccos( -353/7854) #

#B = arccos( 72409/90882) #

#C = arccos(6527/10206) #

Explanation:

Hey people, let's use lower case letters for triangle sides and upper case for the vertices.

These are presumably sides: #a=24.3, b=14.7, c=18.7#. We're after the angles.

Pro Tip: It's generally better to use cosine than sine in a number of places in trig. One reason is that a cosine uniquely determines a triangle angle #(#between #0^circ# and #180^circ),# but the sine is ambiguous; supplementary angles have the same sine. When you have a choice between the Law of Sines and the Law of Cosines, choose cosines.

#c^2 = a^2 + b^2 - 2 a b cos C#

#cos C = {a^2 + b^2 - c^2}/{2 a b}#

#cos C = {24.3^2 + 14.7^2 - 18.7^2}/{2 (24.3)(14.7)} = 6527/10206 #

#cos A = { 14.7^2 + 18.7^2 - 24.3^2 }/{2( 14.7)(18.7)} = -353/7854 #

Negative, an obtuse angle, but small, just a bit more than #90^circ#.

#cos B = { 24.3^2 + 18.7^2 - 14.7^2 }/{2( 24.3)(18.7)} = 72409/90882 #

I hate ruining an exact answer with approximations, so I'll leave the inverse cosine calculator work to you.