Question

If $p_1,p_2,p_3$ are respectively the perpendicular from the vertices of a triangle to the opposite sides, then the value of $\left(\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}\right)$ is equal to

• A. $\frac{1}{r}$
• B. $\frac{1}{R}$
• C. $\frac{1}{\Delta }$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{R}$

In $\Delta ABC$
let $t=\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}=\frac{a\cos A+b\cos B+c\cos C}{2△}$
$\Rightarrow t=\frac{R[\left(2\sin A\cos A\right)+\left(2\sin B\cos B\right)+\left(2\sin C\cos C\right)]}{2△}$ [ from sine rule ]
$\Rightarrow t=\frac{R(\sin 2A+\sin 2B+\sin 2C)}{2△}=\frac{2R(\sin (A+B)\cos (A-B)+\sin C\cos C)}{2△}$
$\Rightarrow t=\frac{2R(\sin (\pi -A-B)\cos (A-B)+\sin C\cos (\pi -A-B))}{2△}=\frac{2R\sin C(\cos (A-B)-\cos (A+B))}{2△}$
$\Rightarrow t=\frac{4R\sin A\sin B\sin C}{2△}$
$\Rightarrow t=\frac{4R\left(a/2R\right)\left(b/2R\right)\left(c/2R\right)}{2△}=\frac{1}{R}\left(\frac{abc}{4R△}\right)=\frac{1}{R}$ [ from sine rule ]
Ans: B