Posted by **can u do it????** on Monday, February 25, 2013 at 11:31am.

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?

- geometry(triangle) -
**Steve**, Monday, February 25, 2013 at 12:05pm
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

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