In a right angled triangle, one acute angle is double the other. The hypotenuse is _____ the smallest side.

Options

A. `sqrt(2)` times
B. three times
C. double
D. 4 times

Answer

To solve the problem, we need to analyze the right-angled triangle where one acute angle is double the other. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Angles**:
- Let one acute angle be \( P \).
- Then, the other acute angle will be \( 2P \).

2. **Use the Triangle Angle Sum Property**:
- In a triangle, the sum of all angles is \( 180^\circ \).
- For our right-angled triangle:
\[ P + 2P + 90^\circ = 180^\circ \]
- Simplifying this gives:
\[ 3P + 90^\circ = 180^\circ \]

3. **Solve for \( P \)**:
- Subtract \( 90^\circ \) from both sides:
\[ 3P = 90^\circ \]
- Dividing both sides by 3:
\[ P = 30^\circ \]

4. **Determine the Angles**:
- The angles in the triangle are:
- \( P = 30^\circ \)
- \( 2P = 60^\circ \)
- Right angle = \( 90^\circ \)

5. **Identify the Sides**:
- In a right-angled triangle, the side opposite the smallest angle is the smallest side.
- Here, the smallest angle is \( 30^\circ \), so the side opposite \( 30^\circ \) (let's call it \( BC \)) is the smallest side.

6. **Using Trigonometric Ratios**:
- We can use the cosine of the angles to find the relationship between the sides.
- For angle \( 60^\circ \):
\[ \cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} \]
- We know \( \cos(60^\circ) = \frac{1}{2} \):
\[ \frac{BC}{AC} = \frac{1}{2} \]

7. **Express the Hypotenuse in Terms of the Smallest Side**:
- Rearranging gives:
\[ AC = 2 \times BC \]
- This means the hypotenuse \( AC \) is twice the smallest side \( BC \).

### Conclusion:
The hypotenuse is **twice** the smallest side.