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In triangle ABC ,angle B=90. P is point in a triangle such that angle APB = angle BPC = angle APC. BP=6,AP=10. FInd PC and area of triangle ABC?

2080 Points
2 years ago
obviously since angle APB = angle BPC = angle APC, they must all be equal to 120 deg.
now, we know that area of a triangle can be expressed as ½ absinC. lets call the area of triangle ABC as A.
then A= sum of areas of triangles APB, BPC, APC
= ½ *6*10*sin120 + ½ *6*x*sin120 + ½ *x*10*sin120, where x= PC.
or A/3= 15+4x........(1)
now, A can also be writen as ½ *AB*BC.
using cosine rule in triangle ABP, we can easily find AB= 14.
so A= 7*BC.
again using cos rule in triangle BPC, we can write BC in terms of x as:
BC = (x^2+6x+36) so that
A= 7(x^2+6x+36) …..(2)
solving (1) and (2), we get
PC= 33, A= 1473
kindly approve :)