Question

In ΔABC, (a + b) cos C + (b + c) cos A + (c + a) cos B is equal to ______.

Options

• a − b + c

• a + b − c

• a + b + c

• a − b − c

Answer 1

In ΔABC, (a + b) cos C + (b + c) cos A + (c + a) cos B is equal to a + b + c.

Explanation:

We know that in a triangle,

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)`

Now, substituting these values into the given equation:

(a + b) cos C + (b + c) cos A + (c + a) cos B

Expanding each term,

`(a + b) xx (a^2 + b^2 - c^2)/(2ab) + (b + c) xx (b^2 + c^2 - a^2)/(2bc) + (c + a) xx (a^2 + c^2 - b^2)/(2ac)`

By simplifying the expressions and using triangle properties, the result simplifies to: 

a + b + c