If in a right angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from opposite vertex, then one of its acute angles is

Options

A. `15^@`
B. `30^@`
C. `45^@`
D. `60^@`

Answer

To solve the problem, we need to find one of the acute angles in a right-angled triangle where the hypotenuse is four times as long as the perpendicular drawn from the opposite vertex to the hypotenuse. Let's denote the following:

- Let \( h \) be the length of the hypotenuse.
- Let \( p \) be the length of the perpendicular drawn from the opposite vertex to the hypotenuse.

According to the problem, we have:
\[ h = 4p \]

### Step-by-step Solution:

1. **Understanding the Triangle**:
- We have a right-angled triangle \( ABC \) where \( \angle C = 90^\circ \).
- Let \( A \) be the vertex opposite the hypotenuse \( BC \).
- The perpendicular from \( A \) to \( BC \) meets \( BC \) at point \( D \).

2. **Using Trigonometric Ratios**:
- In triangle \( ABD \), we can express the tangent of angle \( A \) (denoted as \( \alpha \)):
\[ \tan \alpha = \frac{AD}{BD} \]
- In triangle \( ACD \):
\[ \tan (90^\circ - \alpha) = \frac{AD}{CD} \]

3. **Relating the Sides**:
- From the right triangle properties, we know:
\[ AD = p \quad \text{(the length of the perpendicular)} \]
- The hypotenuse \( h \) can be expressed in terms of \( p \):
\[ h = 4p \]

4. **Using the Area of the Triangle**:
- The area \( A \) of triangle \( ABC \) can be expressed in two ways:
\[ A = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times h \times p \]
- Since \( h = 4p \), we can substitute:
\[ A = \frac{1}{2} \times (4p) \times p = 2p^2 \]

5. **Using the Relationship of Angles**:
- We can use the sine double angle formula:
\[ \sin 2\alpha = 2 \sin \alpha \cos \alpha \]
- Since \( \tan \alpha = \frac{p}{BD} \), we can relate \( \sin \alpha \) and \( \cos \alpha \) to \( p \) and \( h \).

6. **Setting up the Equation**:
- From the relationship of the sides, we can derive:
\[ \sin 2\alpha = \frac{1}{2} \]
- This implies:
\[ 2\alpha = 30^\circ \quad \Rightarrow \quad \alpha = 15^\circ \]

7. **Conclusion**:
- Therefore, one of the acute angles in the triangle is:
\[ \alpha = 15^\circ \]