Question

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

For any general triangle as mentioned above, which of the above is true?

• A. $\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$
• B. $\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}=\frac{a^2+b^2+c^2}{2abc}$
• C. $\frac{sinA}{a}+\frac{sinB}{b}+\frac{sinC}{c}=\frac{3}{2R}$
• D. $\frac{sin2A}{a^2}=\frac{sin2B}{b^2}=\frac{sin2C}{c^2}$

Answer 1

Answer: (B), (C)

$\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}=\frac{a^2+b^2+c^2}{2abc}$
$\frac{sinA}{a}+\frac{sinB}{b}+\frac{sinC}{c}=\frac{3}{2R}$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$

Hence

$\frac{\sin A}{a}+\frac{\sin B}{b}+\frac{\sin C}{c}=\frac{3}{2R}$

Now

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$

Or

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Or

$\frac{2\sin A.cosA}{2acosA}=\frac{2\sin B.cosB}{2bcosB}=\frac{2\sin C.cosC}{2ccosC}$

Or

$\frac{\sin 2A}{2a.cosA}=\frac{\sin 2B}{2bcosB}=\frac{\sin 2C}{2ccosC}$.

Hence A and D are false.

Now

$\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$

$=\frac{1}{2abc}\left[\right(b^2+c^2-a^2)+(a^2+c^2-b^2)+(a^2+b^2-c^2\left)\right]$

$=\frac{1}{2abc}\left[2\right(a^2+b^2+c^2)-a^2+b^2+c^2]$

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