In acute angled triangle `A B C ,A D` is the altitude. Circle drawn with `A D` as its diameter cuts `A Ba n dA Ca tPa n dQ ,` respectively. Length of `P Q` is equal to `/(2R)` (b) `(a b c)/(4R^2)` `2RsinAsinBsinC` (d) Δ`/R`

Explanation

To solve the problem, we need to find the length of segment \( PQ \) in triangle \( ABC \) where \( AD \) is the altitude from vertex \( A \) to side \( BC \). The circle with diameter \( AD \) intersects \( AB \) at point \( P \) and \( AC \) at point \( Q \).

### Step-by-Step Solution:

1. **Identify the Triangle and Points**:
We have triangle \( ABC \) with altitude \( AD \). Points \( P \) and \( Q \) are where the circle with diameter \( AD \) intersects \( AB \) and \( AC \), respectively.

2. **Use the Sine Rule in Triangle \( APQ \)**:
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