If in a `DeltaABC`, the altitudes from the vertices A, B, C on opposite sides are in H.P, then sin A, sin B, sin C are in

Options

A. HP
B. AGP
C. AP
D. GP

Answer

To solve the problem, we need to show that if the altitudes from the vertices A, B, and C of triangle ABC onto the opposite sides are in Harmonic Progression (HP), then the sines of the angles A, B, and C are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
Let the altitudes from vertices A, B, and C to the opposite sides BC, CA, and AB be denoted as AD, BE, and CF respectively. We know that AD, BE, and CF are in HP.

2. **Using the Definition of HP**:
If three quantities are in HP, then their reciprocals are in AP. Therefore, we can write:
\[ \frac{1}{AD}, \frac{1}{BE}, \frac{1}{CF} \text{ are in AP.} \]

3. **Relating Altitudes to Sines**:
The altitudes can be expressed in terms of the sides of the triangle and the sine of the angles:
- \( AD = \frac{BC \cdot \sin A}{AB} \)
- \( BE = \frac{CA \cdot \sin B}{BC} \)
- \( CF = \frac{AB \cdot \sin C}{CA} \)

However, we can simplify this by noting that:
\[ AD = \frac{2 \times \text{Area of } \triangle ABC}{BC}, \quad BE = \frac{2 \times \text{Area of } \triangle ABC}{CA}, \quad CF = \frac{2 \times \text{Area of } \triangle ABC}{AB} \]

4. **Expressing the Reciprocals**:
Thus, we can write the reciprocals:
\[ \frac{1}{AD} = \frac{BC}{2 \times \text{Area of } \triangle ABC}, \quad \frac{1}{BE} = \frac{CA}{2 \times \text{Area of } \triangle ABC}, \quad \frac{1}{CF} = \frac{AB}{2 \times \text{Area of } \triangle ABC} \]

5. **Setting Up the AP Condition**:
Now, since \( \frac{1}{AD}, \frac{1}{BE}, \frac{1}{CF} \) are in AP, we can write:
\[ 2 \cdot \frac{CA}{2 \times \text{Area of } \triangle ABC} = \frac{BC}{2 \times \text{Area of } \triangle ABC} + \frac{AB}{2 \times \text{Area of } \triangle ABC} \]
Simplifying this gives:
\[ 2CA = BC + AB \]

6. **Relating to Sines**:
From the relationships of the sides and angles in a triangle, we know:
\[ \frac{BC}{\sin A} = \frac{CA}{\sin B} = \frac{AB}{\sin C} = 2R \text{ (circumradius)} \]
Thus, we can express:
\[ \frac{1}{\sin A}, \frac{1}{\sin B}, \frac{1}{\sin C} \text{ are in AP.} \]

7. **Conclusion**:
Therefore, since the reciprocals of the sines are in AP, it follows that:
\[ \sin A, \sin B, \sin C \text{ are in AP.} \]

### Final Answer:
Thus, if the altitudes from the vertices A, B, C of triangle ABC onto the opposite sides are in HP, then \( \sin A, \sin B, \sin C \) are in AP.