Question

If $p_1,p_2,p_3$ are the perpendiculars drawn from the vertices of a triangle $ABC$ to the opposite sides respectively, then find the value of

$\frac{cosA}{p_1}+\frac{cosB}{p_2}+\frac{cosC}{p_3}$

• A. $\frac{1}{r}$
• B. $\frac{1}{R}$
• C. $\frac{1}{\Delta }$
• D. None

Answer 1

Answer: (B)

$\frac{1}{R}$

We have, $\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$

$[Area=\frac{1}{2}\times base\times altitude]$

$\therefore \frac{1}{p_1}=\frac{a}{2\Delta },\frac{1}{p_2}=\frac{b}{2\Delta },\frac{1}{p_3}=\frac{c}{2\Delta }$

$\frac{cosA}{p_1}+\frac{cosB}{p_2}+\frac{cosC}{p_3}=\frac{1}{2\Delta }(acosA+bcosB+ccosC)$

$=\frac{R}{\Delta }(sinAcosA+sinBsinB+sinCsinC)$ $\left[fromsinerule\right]$

$=\frac{R}{2\Delta }(sin2A+sin2B+sin2C)$

$=R\frac{4sinAsinBsinC}{2\Delta }=\frac{2RsinAsinBsinC}{\Delta }$ $\dots \left[\text{from conditional identity}\right]$

$=\frac{2R}{\Delta }\frac{2\Delta }{bc}\times \frac{2\Delta }{ca}\times \frac{2\Delta }{ab}$

$=\frac{16R{\Delta }^2}{a^2{b}^2{c}^2}$

$=\frac{16R{\Delta }^2}{(4R\Delta {)}^2}$

$=\frac{1}{R}$

$\therefore \frac{cosA}{p_1}+\frac{cosB}{p_2}+\frac{cosC}{p_3}=\frac{1}{R}$