Grade 12th passAnalytical Geometry

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?

Jitender Saini

6 Years agoGrade 12th pass

1 Answer

Aditya Gupta

Approved Tutor Answer6 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= 147√3

kindly approve :)

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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?

Jitender Saini

1 Answer

Aditya Gupta

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