Triangle ABC is a right triangle with ∠ABC=90∘. P is a point within triangle ABC such that ∠APB=∠BPC=∠CPA=120∘. If PA=15 and PB=6, what is the value of PC?

Answer 1

There's probably a good geometric way to do it, but here's a trig way, using the law of cosines.

AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351

AC^2 = PC^2 + 15^2 - 2*PC*15(-1/2)
BC^2 = PC^2 + 6^2 - 2*PC*6(-1/2)

AC^2-BC^2 = AB^2 = 351, so

351 = 189 + 9PC
9PC = 162