Question

PLEASE SOLVE THIS...

Answer 1

(1).

Given Equation is a^4 + b^4 + c^4 = 2a^2b^2 + b^2c^2 + 2c^2a^2

It can be written as:

= > a^4 + b^4 + c^4 - 2a^2b^2 - 2c^2a^2 + (2b^2c^2 - b^2c^2) = 0

= > a^4 + b^4 + c^4 - 2a^2b^2 + 2b^2c^2 -  2c^2a^2 = b^2c^2

= > (a^2)^2 + (b^2)^2 + (c^2)^2 - 2a^2b^2 + 2b^2c^2 - 2c^2a^2 = b^2c^2

It is in the form of a + b + c - 2ab + 2bc - 2ca = (a  b - c)^2

= > (a^2 - b^2 - c^2)^2 = b^2c^2

= > a^2 - b^2 - c^2 = bc

= > b^2 + c^2 - a^2 = bc

= > b^2 + c^2 - a^2/bc = 1

Divide by '2' on both sides, we get

= > (b^2 + c^2 - a^2)/2bc = 1/2

We know that cosA = (b^2 + c^2 - a^2)/2bc.

= > cosA = 1/2

= > cosA = 60

= > A = 60.

Given,

⇒ sinA

⇒ sin(60)

⇒ √3/2.

Therefore, the answer is Option (B) - √3/2.

Note: The answer will be same for sin(120) also. Because here I have considered cosA as positive.

Hope this helps!