Question

If $P_1{P}_2,P_3$ are the perpendiculars of a triangle $ABC$ from the vertices $A,B,C$ to sides opposites to $\angle A,\angle B,\angle C$ respectively and $2S$ be the perimeter of triangle ABC $,\Delta$ is the area of the triangle.

If in a triangle ABC, sin A, sin B, sin C are in A.P., then

• A. $P_1,P_2,P_3$,are in A.P.
• B. $P_1,P_2,P_3$,are in H.P.
• C. $P_1,P_2,P_3$,are in G.P.
• D. None of these

Answer 1

The correct option is B $P_1,P_2,P_3$,are in H.P.

We have,

$AD=b\sin C$
$BE=c\sin A$
$CF=a\sin B$.

Now,
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$a=ksinA$, $b=ksinB$, $c=ksinC$

Hence
$AD=k\sin B\sin C$

$BE=k\sin A\sin C$

$CF=k\sin A\sin B$.

Now,

$2\sin B=\sin C+\sin A$

$\frac{1}{AD}+\frac{1}{CF}$
$=\frac{1}{k}[\frac{1}{\sin B.\sin C}+\frac{1}{\sin A.\sin B}]$
$=\frac{1}{k.\sin B}[\frac{1}{\sin C}+\frac{1}{\sin A}]$
$=\frac{1}{k.\sin B}\left[\frac{\sin A+\sin C}{\sin C.\sin A}\right]$
$=\frac{1}{k\sin B}\left[\frac{2\sin B}{\sin C.\sin A}\right]$
$=\frac{2}{k.\sin A.\sin C}$
$=\frac{2}{BE}$

Hence they are in H.P.