First slide

Properties of triangles

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

Moderate

A

$1 / r$
B

$1 / R$
C

$1 / Δ$
D

none of these

Solution

We have,
$\begin{array}{l} \frac{\cos A}{p_{1}} + \frac{\cos B}{p_{2}} + \frac{\cos C}{p_{3}} \\ = \frac{1}{2 Δ} (a \cos A + b \cos B + i \cos C) \end{array}$
$\begin{array}{l} = \frac{R}{Δ} (\sin A \cos A + \sin B \cos B + \sin C \cos C) \\ = \frac{R}{2 Δ} (\sin 2 A + \sin 2 B + \sin 2 C) \\ = R \frac{4 \sin A \sin B \sin C}{2 Δ} = \frac{2 R \sin A \sin B \sin C}{Δ} \\ = \frac{2 R}{Δ} \times \frac{2 Δ}{b c} \times \frac{2 Δ}{c a} \times \frac{2 Δ}{a b} = \frac{16 R Δ^{2}}{a^{2} b^{2} c^{2}} = \frac{16 R Δ^{2}}{(4 R Δ)^{2}} = \frac{1}{R} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App