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. **Define the Triangle and Variables**:
Let \( \triangle ABC \) be a right-angled triangle with \( \angle B = 90^\circ \). Let \( AC \) be the hypotenuse, \( AB \) be the opposite side, and \( BC \) be the adjacent side. Let the perpendicular from vertex \( B \) to \( AC \) be \( P \).

2. **Express the Hypotenuse**:
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