Question

If in a $△ABC$, $\sin A,\sin B,\sin C$ are in AP then

• A. the altitudes are in AP
• B. the altitudes are in HP
• C. the angles are in AP
• D. the angles are in HP

Answer 1

The correct option is B the altitudes are in HP

If $p_1,p_2$ and $p_3$ be the altitudes of a triangle, then $\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$
$\Rightarrow a=\frac{2\Delta }{p_1},b=\frac{2\Delta }{p_2},c=\frac{2\Delta }{p_3}$
Since, $\sin A,\sin B$ and $\sin C$ are in AP.
$\sin B=\frac{\sin A+\sin C}{2}$
$\Rightarrow b=\frac{a+c}{2}$
$\Rightarrow \frac{2\Delta }{p_2}=\frac{\frac{2\Delta }{p_1}+\frac{2\Delta }{p_3}}{2}$
$\Rightarrow \frac{2}{p_2}=\frac{p_3+p_1}{p_1{p}_3}$
$\Rightarrow p_2=\frac{2p_1{p}_3}{p_1+p_3}$
Hence, altitudes are in H.P.