Question

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

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

Answer 1

Answer: (B)

We have, $\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$
$\therefore \frac{1}{p_1}=\frac{a}{2\Delta },\frac{1}{p_2}=\frac{b}{2\Delta },\frac{1}{p_3}=\frac{c}{2\Delta }$
$\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}=\frac{1}{2\Delta }(a\cos A+b\cos B+c\cos C)$
$=\frac{R}{\Delta }(\sin A\cos A+\sin B\cos B+\sin C\cos C)$
$=\frac{R}{2\Delta }(\sin 2A+\sin 2B+\sin 2C)$
$=R\frac{4\sin A\sin B\sin C}{2\Delta }$
$=\frac{2R}{\left(\frac{abc}{4R}\right)}\times \frac{a}{2R}\times \frac{b}{2R}\times \frac{c}{2R}=\frac{1}{R}$