In a right angled triangle the hypotenuse is `2sqrt(2)` times the length of perpendicular drawn from the opposite vertex, on the hypertenuse, then the other two angles, are

Options

A. `(pi)/(3), (pi)/(6)`
B. `(pi)/(4), (pi)/(4)`
C. `(pi)/(8), (3pi)/(8)`
D. `(pi)/(12), (5pi)/(12)`

Answer

To solve the problem, we need to analyze the right-angled triangle and use trigonometric identities to find the angles based on the given condition.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Let the right-angled triangle be \( ABC \) with \( \angle C = 90^\circ \). Let \( AB \) be the hypotenuse, \( AC \) be the base, and \( BC \) be the height (perpendicular drawn from vertex \( C \) to hypotenuse \( AB \)).

2. **Given Condition**:
We are given that the hypotenuse \( AB = 2\sqrt{2} \times BC \). Let \( BC = h \). Therefore, we can express the hypotenuse as:
\[ AB = 2\sqrt{2}h \]

3. **Using the Area of the Triangle**:
The area \( A \) of triangle \( ABC \) can be expressed in two ways:
- Using the base and height:
\[ A = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times AC \times h \]
- Using the hypotenuse and height:
\[ A = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times (2\sqrt{2}h) \times h = \sqrt{2}h^2 \]

4. **Equating the Areas**:
Setting the two expressions for the area equal gives:
\[ \frac{1}{2} \times AC \times h = \sqrt{2}h^2 \]
Dividing both sides by \( h \) (assuming \( h \neq 0 \)):
\[ \frac{1}{2} \times AC = \sqrt{2}h \]
Thus, we can express \( AC \) as:
\[ AC = 2\sqrt{2}h \]

5. **Finding Angles Using Trigonometric Ratios**:
We can use the sine and cosine ratios to find the angles \( A \) and \( B \):
- For angle \( A \):
\[ \sin A = \frac{BC}{AB} = \frac{h}{2\sqrt{2}h} = \frac{1}{2\sqrt{2}} \quad \Rightarrow \quad A = \sin^{-1}\left(\frac{1}{2\sqrt{2}}\right) \]
- For angle \( B \):
\[ \sin B = \frac{AC}{AB} = \frac{2\sqrt{2}h}{2\sqrt{2}h} = 1 \quad \Rightarrow \quad B = 90^\circ \]

6. **Finding the Other Angle**:
Since the sum of angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \quad \Rightarrow \quad A + 90^\circ + 0^\circ = 180^\circ \quad \Rightarrow \quad A = 90^\circ - B \]

7. **Final Angles**:
The angles in triangle \( ABC \) are:
- \( A = 45^\circ \) (or \( \frac{\pi}{4} \))
- \( B = 90^\circ \) (or \( \frac{\pi}{2} \))
- \( C = 45^\circ \) (or \( \frac{\pi}{4} \))

### Final Answer:
The other two angles in the triangle are \( 45^\circ \) and \( 45^\circ \).