Question

how can i solve this ..anyone?​

Answer 1

In a triangle ABC, it's given that,

$\longrightarrow \cos\left(\dfrac{A}{2}\right)=\sqrt{\dfrac{b+c}{2c}}$

Squaring,

$\longrightarrow \cos^2\left(\dfrac{A}{2}\right)=\dfrac{b+c}{2c}$

Multiplying by 2,

$\longrightarrow2\cos^2\left(\dfrac{A}{2}\right)=\dfrac{b+c}{c}$

Subtract 1 from both sides,

$\longrightarrow2\cos^2\left(\dfrac{A}{2}\right)-1=\dfrac{b+c}{c}-1$

Since $2\cos^2x-1=\cos(2x),$

$\longrightarrow\cos A=\dfrac{b}{c}$

This implies ABC is a right triangle of sides a, b and c, with hypotenuse c and perpendicular sides a and b.

Because cosine of an angle (A) is adjacent side (b) divided by hypotenuse (c).

Therefore, by Pythagoras' Theorem,

$\longrightarrow\underline{\underline{a^2+b^2=c^2}}$

Hence (a) is the answer.