Step-by-step Solution:

Concept:

Area of \( \triangle ABC = A = \frac{1}{2} \times \text{Base} \times \text{Height} \)

Let \( a \), \( b \), and \( c \) be the sides of the triangle.

If \( \frac{1}{a} \), \( \frac{1}{b} \), and \( \frac{1}{c} \) are in HP, then \( a \), \( b \), and \( c \) are in AP.

If \( a \), \( b \), and \( c \) are in AP, then \( 2b = a + c \).

The Law of Sines says that in any given triangle, the ratio of any side length to the sine of its opposite angle is the same for all three sides of the triangle.

\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \)

 

Calculations:

Given, in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP

 

 

In triangle ABC, the altitudes from the vertices A to BC, B to AC, and C to AB.

\( \Rightarrow AD, BE, \text{ and } CF \text{ are the altitudes and they are in HP.} \)

\( \Rightarrow \text{Area of } \triangle ABC = A = \frac{1}{2} \times AD \times BC \)

\( \Rightarrow AD = \frac{2A}{a} \)

Similarly, \( BE = \frac{2A}{b} \) and \( CF = \frac{2A}{c} \)

\( \Rightarrow \frac{2A}{a}, \frac{2A}{b}, \text{ and } \frac{2A}{c} \text{ are in HP.} \)

\( \Rightarrow \frac{1}{a}, \frac{1}{b}, \text{ and } \frac{1}{c} \text{ are in HP.} \)

\( \Rightarrow a, b, \text{ and } c \text{ are in AP.} \)

\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \) (sine rule)

\( \Rightarrow a = k \sin A, b = k \sin B, \text{ and } c = k \sin C \)

As we know, If \( a \), \( b \), and \( c \) are in AP then \( 2b = a + c \)

\( \Rightarrow 2k \sin B = k \sin A + k \sin C \)

\( \Rightarrow 2 \sin B = \sin A + \sin C \)

Hence \( \sin A, \sin B, \sin C \) are in AP