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, $\Delta=\frac{1}{2} a p_{1}=\frac{1}{2} b p_{2}=\frac{1}{2} c p_{3}$
$\therefore \frac{1}{p_{1}}=\frac{a}{2 \Delta}, \frac{1}{p_{2}}=\frac{b}{2 \Delta}, \frac{1}{p_{3}}=\frac{c}{2 \Delta}$
$\frac{\cos A}{p_{1}}+\frac{\cos B}{p_{2}}+\frac{\cos C}{p_{3}}=\frac{1}{2 \Delta}(a \cos A+b \cos B+c \cos C)$
$=\frac{ R }{\Delta}(\sin A \cos A +\sin B \cos B +\sin C \cos C )$
$=\frac{ R }{2 \Delta}(\sin 2 A +\sin 2 B +\sin 2 C )$
$= R \frac{4 \sin A \sin B \sin C }{2 \Delta} $
$=\frac{2 R }{\left(\frac{ abc }{4 R }\right)} \times \frac{ a }{2 R } \times \frac{ b }{2 R } \times \frac{ c }{2 R }=\frac{1}{ R }$