In triangle $ABC$, $AB = BC = 25$ and $AC = 30$. What is $\sin \angle ABC$?
We have something like this
B
25 25
A 30 C
Using the Law of Cosines
30^2 = 25^2 + 25^2 - 2(25*25) cos ABC
cos (ABC) = [ 30^2 - 25^2 - 25^2 ] / [ [ -2 (25 * 25) ]
cos (ABC) = 7/25
So..... sin (ABC) = sqrt (25^2 - 7^2] / 25 = sqrt [ 625 - 49 ] /25 =
sqrt [ 576 ] / 25 =
24 / 25