In a right angled triangle, the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex. One of the acute angle is

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle**:
Let's denote the right-angled triangle as \( \triangle ABC \) where \( \angle C = 90^\circ \). Let \( AM \) be the perpendicular drawn from vertex \( A \) to the hypotenuse \( BC \).

2. **Given Relation**:
We know that the hypotenuse \( BC \) is four times the length of the perpendicular \( AM \). Therefore, we can write:
\[ BC = 4 \cdot AM \]

3. **Assigning Angles**:
Let \( \angle A = \alpha \) and \( \angle B = 90^\circ - \alpha \).

4. **Using Trigonometric Ratios**:
In triangle \( \triangle ABM \):
- The tangent of angle \( \alpha \) is given by:
\[ \tan \alpha = \frac{AM}{MB} \]
- Rearranging gives:
\[ MB = \frac{AM}{\tan \alpha} \]

5. **In triangle \( \triangle ACM \)**:
- The tangent of angle \( 90^\circ - \alpha \) (which is \( \cot \alpha \)) is given by:
\[ \tan(90^\circ - \alpha) = \frac{AM}{MC} \]
- Rearranging gives:
\[ MC = \frac{AM}{\cot \alpha} = AM \cdot \tan \alpha \]

6. **Finding the Length of the Hypotenuse**:
The length of the hypotenuse \( BC \) can be expressed as:
\[ BC = MB + MC = \frac{AM}{\tan \alpha} + AM \cdot \tan \alpha \]
Factoring out \( AM \):
\[ BC = AM \left( \frac{1}{\tan \alpha} + \tan \alpha \right) \]

7. **Substituting the Given Relation**:
Since \( BC = 4 \cdot AM \), we equate:
\[ 4 \cdot AM = AM \left( \frac{1}{\tan \alpha} + \tan \alpha \right) \]
Dividing both sides by \( AM \) (assuming \( AM \neq 0 \)):
\[ 4 = \frac{1}{\tan \alpha} + \tan \alpha \]

8. **Letting \( x = \tan \alpha \)**:
Thus we have:
\[ 4 = \frac{1}{x} + x \]
Multiplying through by \( x \):
\[ 4x = 1 + x^2 \]
Rearranging gives:
\[ x^2 - 4x + 1 = 0 \]

9. **Solving the Quadratic Equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]
Thus, \( \tan \alpha = 2 + \sqrt{3} \) (we take the positive root since angles are positive).

10. **Finding the Angle**:
To find \( \alpha \), we can use the inverse tangent function:
\[ \alpha = \tan^{-1}(2 + \sqrt{3}) \]

11. **Calculating the Angle**:
Using a calculator or trigonometric tables, we find:
\[ \alpha \approx 15^\circ \]

### Final Answer:
One of the acute angles in the triangle is \( 15^\circ \).

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