Question

If in a $△ABC$, the altitudes from the vertices $A,B,C$ on the opposite side are in HP, then $\sin A,\sin B,$ and $\sin C$ are in

• A. HP
• B. Arithmetic – Geometric Progression
• C. AP
• D. GP

Answer 1

The correct option is C

AP

Explanation for the correct option:

Find the progression formed by $\sin A,\sin B,\sin C$.

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

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

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

Again as $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in HP, so $a,b,c$ are in AP.

Now, as $a,b,c$ are in AP, so using the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ it can be said that $\sin A,\sin B,\sin C$ are also in AP.

Hence, the correct option is C.