Question

In any triangle ABC, prove the following :cosAa+cosBb+cosCc=a2+b2+c2 2abc

Answer 1

Simple amplification of  Cosine law.

Step-by-step explanation:

To solve this we will be using cosine law of triangles.

By cosine law of triangles we have:

c^2 = a^2 + b^2 − 2ab cos(C)

cos(C) = (c^2 - a^2 - b^2) / (-2ab) -->

--> cos(C) = (a^2 + b^2 - c^2)/2ab

-Okay so we will use this for each angle respectively, and we get:

cos(A) = (b^2 + c^2 - a^2)/2bc  

cos(B) = (a^2 + c^2 - b^2)/2ac  

cos(C) = (a^2 + b^2 - c^2)/2ab  

cos(A)/a + cos(B)/b + cos(C)/c = (b^2 + c^2 - a^2)/2abc + (a^2 + c^2 - b^2)/2abc + (a^2 + b^2 - c^2)/2abc   = (b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2)/2abc = (a^2 + b^2 + c^2)/2abc .