Question

In any triangle abc prove that cosa/a + cosb/b + cosc/c = (a^2+b^2+c^2)/2abc

Answer 1

Answer:

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 .

Answer 2

we know from properties of triangle.

$cosA=\frac{b^2+c^2-a^2}{2bc}\\\\cosB=\frac{c^2+a^2-b^2}{2ca}\\\\cosC=\frac{a^2+b^2-c^2}{2ab}$

$LHS=\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}\\\\=\frac{\frac{b^2+c^2-a^2}{2bc}}{a}+\frac{\frac{c^2+a^2-b^2}{2ca}}{b}+\frac{\frac{a^2+b^2-c^2}{2ab}}{c}\\\\=\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}\\\\=\frac{(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)}{2abc}\\\\=\frac{a^2+b^2+c^2}{2abc}=RHS$