Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively,

then $\frac{b\cos A+a\cos B}{\cos C}$ is equal to

• A. $\frac{2abc}{a^2+b^2+c^2}$
• B. $\frac{abc}{b^2+c^2-a^2}$
• C. $\frac{abc}{b^2+c^2+a^2}$
• D. $\frac{2abc}{b^2+a^2-c^2}$

Answer 1

The correct option is D $\frac{2abc}{b^2+a^2-c^2}$

Given a $\Delta$ ABC whose sides are a,b,c & angles are A,B,C

By $\text{cosine formula}$,

$\cos \left(C\right)=\frac{b^2+a^2-c^2}{2ba}$

By $\text{Projection formula}$,

$(b\times \cos A+a\times \cos B)=c$

Therefore, $\frac{b\left(\cos A\right)+a\left(\cos B\right)}{\cos C}=\frac{c}{\cos C}$

$=\frac{c\times \left(2ab\right)}{b^2+a^2-c^2}$

$=\frac{2abc}{b^2+a^2-c^2}$

hence, option (D) is correct.