In a right angled triangle the hypotenuse is `2sqrt(2)` times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are (a)`pi/3a n dpi/6` (b) `pi/8a n d(3pi)/8` (c) `pi/4a n dpi/4` (d) `pi/5a n d(3pi)/(10)`

Explanation

To solve the problem, we need to find the other acute angles of a right-angled triangle where the hypotenuse is \(2\sqrt{2}\) times the perpendicular drawn from the opposite vertex.

### Step-by-step Solution:

1. **Understanding the Triangle**:
Let triangle ABC be a right-angled triangle at vertex B. Let BM be the perpendicular from vertex B to side AC. We denote the length of BM as \(p\). According to the problem, the hypotenuse AC is given by:
\[
AC = 2\sqrt{2} \cdot BM = 2\sqrt{2} \cdot p
...