if you have a triangle that has sides with length of 14, 19, and 12, how could you get the angle using cosines
rearrange a b Ánd c fór which ever Angle you want a is always the side opposite the Angle Ánd b Ánd c are the other two
+129845 Using the Law of Cosines, we can find the angle opposite the longest side thusly :
19^2 = 14^2 + 12^2 - 2(14)(12)cosθ where θ is the angle we're looking for
Rearrange as
[19^2 - 14^2 - 12^2] / [ -2(14)(12)] = cosθ
Using the cosine inverse [arccos], we can find θ
θ = arcos ( [19^2 - 14^2 - 12^2] / [ -2(14)(12)] ) = about 93.58°
Now.......we can can find a second angle, call it B, using the Law of Sines
sinB / 14 = sin (93.58) / 19 multiply both sides by 14
sin B = [14 sin(93.58)/ 19]
And using the sine inverse [arcsin] we can find B
B = arcsin [14 sin(93.58)/ 19] = about 47.34°
And the remaining angle = [180 -93.58 - 47.34]° = about 39.08°
