Question

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

• A. the radius 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

Explanation for the correct option.

Find the correct relation.

As $\sin A,\sin B,\sin C$ are in AP, so for some constant $k$, $k\sin A,k\sin B,k\sin C$ are also in AP.

Now by sine rule: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$.

So as $k\sin A,k\sin B,k\sin C$ are in AP, so $a,b,c$ are also in AP.

Now the area of the triangle is given as: $∆=\frac{1}{2}{h}_{1}a=\frac{1}{2}{h}_{2}b=\frac{1}{2}{h}_{3}c$.

So the ratio $h_1:h_2:h_3$ is given as

$\begin{array}{rcl}{h}_1:h_2:h_3& =& \frac{2∆}{a}:\frac{2∆}{b}:\frac{2∆}{c}\\ & =& \frac{1}{a}:\frac{1}{b}:\frac{1}{c}\end{array}$

As $a,b,c$ are in AP, so $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in HP and thus the altitudes $h_1,h_2,h_3$ are in HP.

Hence, the correct option is B.