If in a triangle `PQR; sin P, sin Q, sin R` are in A.P; then (A)the altitudes are in AP (B)the altitudes are in HP (C)the altitudes are in GP (D)the medians are in AP

Options

A. the altitudes are in AP
B. the altitudes are in HP
C. the medians are in GP
D. the medians are in AP

Explanation

To solve the problem, we need to analyze the relationship between the angles of triangle \( PQR \) and their corresponding altitudes based on the given condition that \( \sin P, \sin Q, \sin R \) are in arithmetic progression (A.P).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( \sin P, \sin Q, \sin R \) are in A.P. This means that:
\[ 2 \sin Q = \sin P + \sin R \]

2. **Using the Area of the Triangle**:
The area \( \Delta \) of triangle \( PQR \) can be expressed in terms of the sides and the corresponding altitudes:
\[ \Delta = \frac{1}{2} \times a \times h_P = \frac{1}{2} \times b \times h_Q = \frac{1}{2} \times c \times h_R \]
where \( a, b, c \) are the lengths of sides opposite to angles \( P, Q, R \) respectively, and \( h_P, h_Q, h_R \) are the altitudes from vertices \( P, Q, R \).

3. **Finding the Altitudes**:
From the area formulas, we can express the altitudes in terms of the area and the sides:
\[ h_P = \frac{2\Delta}{a}, \quad h_Q = \frac{2\Delta}{b}, \quad h_R = \frac{2\Delta}{c} \]

4. **Relating the Sides to the Sines**:
From the Law of Sines, we know:
\[ \frac{a}{\sin P} = \frac{b}{\sin Q} = \frac{c}{\sin R} = 2R \]
where \( R \) is the circumradius of triangle \( PQR \). Therefore, we can express the sides in terms of the sines:
\[ a = 2R \sin P, \quad b = 2R \sin Q, \quad c = 2R \sin R \]

5. **Substituting into Altitude Formulas**:
Substituting these expressions for \( a, b, c \) into the formulas for the altitudes:
\[ h_P = \frac{2\Delta}{2R \sin P} = \frac{\Delta}{R \sin P}, \quad h_Q = \frac{2\Delta}{2R \sin Q} = \frac{\Delta}{R \sin Q}, \quad h_R = \frac{2\Delta}{2R \sin R} = \frac{\Delta}{R \sin R} \]

6. **Establishing the Relationship**:
Since \( \sin P, \sin Q, \sin R \) are in A.P., we can deduce that \( \frac{1}{\sin P}, \frac{1}{\sin Q}, \frac{1}{\sin R} \) will be in Harmonic Progression (H.P.). This is because if the sines are in A.P., their reciprocals will be in H.P.

7. **Conclusion**:
Since the altitudes \( h_P, h_Q, h_R \) are proportional to \( \frac{1}{\sin P}, \frac{1}{\sin Q}, \frac{1}{\sin R} \), we conclude that the altitudes are in H.P.

Thus, the correct answer is:
**(B) the altitudes are in HP.**