Question

For a triangle $ABC,$ which of the following is true?

• A. $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$
• B. $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}$
• C. $\frac{\sin A}{a}+\frac{\sin B}{b}+\frac{\sin C}{c}=\frac{3}{2R}$
• D. $\frac{\sin 2A}{a^2}=\frac{\sin 2B}{b^2}=\frac{\sin 2C}{c^2}$

Answer 1

Answer: (B), (C)

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

Option$\left(a\right)$
$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$
$\Rightarrow \frac{\frac{b^2+c^2-a^2}{2bc}}{a}=\frac{\frac{c^2+a^2-b^2}{2ca}}{b}=\frac{\frac{a^2+b^2-c^2}{2ab}}{c}$
$\Rightarrow \frac{b^2+c^2-a^2}{2abc}=\frac{c^2+a^2-b^2}{2abc}=\frac{a^2+b^2-c^2}{2abc}$
$\Rightarrow b^2+c^2-a^2=c^2+a^2-b^2=a^2+b^2-c^2$
$\therefore$ It is possible when $a,b,c$ are all zero, which is impossible (since sides$\ne 0$)
Option$\left(b\right)$
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$
$=\frac{\frac{b^2+c^2-a^2}{2bc}}{a}+\frac{\frac{c^2+a^2-b^2}{2ca}}{b}+\frac{\frac{a^2+b^2-c^2}{2ab}}{c}$
$=\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$
$=\frac{a^2+b^2+c^2}{2abc}$
Option$\left(c\right)$
$\frac{\sin A}{a}+\frac{\sin B}{b}+\frac{\sin C}{c}=\frac{1}{2R}+\frac{1}{2R}+\frac{1}{2R}=\frac{3}{2R}$

Option$\left(d\right)$

$\because \frac{\sin 2A}{a^2}=\frac{2\sin A\cos A}{4R^2{\sin }^{2}A}$
$=\frac{\cos A}{2R^2\sin A}$

$=\frac{\cot A}{2R^2}$
Similarly, we have $\frac{\sin 2B}{b^2}=\frac{\cot B}{2R^2}$
and $\frac{\sin 2C}{c^2}=\frac{\cot C}{2R^2}$
Hence $\frac{\cot A}{2R^2}=\frac{\cot B}{2R^2}=\frac{\cot C}{2R^2}$
$\Rightarrow \cot A=\cot B=\cot C$ is possible when $A=B=C$. which is always not true.