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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?`
9 months 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 thatA= 7√(x^2+6x+36) …..(2)solving (1) and (2), we getPC= 33, A= 147√3kindly approve :)
```
9 months ago
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