ABC is a triangle right angled at B. The hypotenuse (AC) is four times the perpendicular (BD) drawn to it from the opposite vertex and `ADltDC`.
What is one of the acute angle of the triangle?

Options

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

Answer

To solve the problem step by step, we will analyze the given information about triangle ABC, which is right-angled at B.

### Step 1: Understand the Triangle
Given that triangle ABC is right-angled at B, we know:
- AC is the hypotenuse.
- BD is the perpendicular from vertex B to hypotenuse AC.

### Step 2: Set Up the Relationship
According to the problem, the hypotenuse AC is four times the length of the perpendicular BD. We can express this relationship mathematically:
- Let BD = m. Then, AC = 4m.

### Step 3: Use Trigonometric Ratios
In triangle ABC, we can use the cosine and sine functions to relate the sides:
1. **Cosine of angle A**:
\[ \cos(\theta) = \frac{BC}{AC} = \frac{BC}{4m} \]
(where θ is angle A)

2. **Sine of angle C**:
\[ \sin(\theta) = \frac{BD}{AC} = \frac{m}{4m} = \frac{1}{4} \]

### Step 4: Relate the Sine and Cosine
We can multiply the two equations derived from the trigonometric ratios:
\[ \cos(\theta) \cdot \sin(\theta) = \frac{BC}{4m} \cdot \frac{m}{4m} = \frac{BC}{16m} \]

### Step 5: Solve for BC
From the sine function:
\[ \sin(\theta) = \frac{1}{4} \implies \theta = \sin^{-1}\left(\frac{1}{4}\right) \]
Now, using the identity \(2 \cos(\theta) \sin(\theta) = \sin(2\theta)\):
\[ \sin(2\theta) = \frac{1}{8} \]

### Step 6: Find the Angle
Using the inverse sine function:
\[ 2\theta = \sin^{-1}\left(\frac{1}{8}\right) \]
Thus,
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]

### Step 7: Calculate the Angles in Triangle ABC
Since triangle ABC is a right triangle:
\[ \angle A + \angle B + \angle C = 180^\circ \]
Given that angle B is 90 degrees, we have:
\[ \angle A + \angle C = 90^\circ \]

### Step 8: Determine One of the Acute Angles
Using the relationship from the sine and cosine:
- If we find that \(\theta\) corresponds to \(15^\circ\), then:
\[ \angle A = 75^\circ \quad \text{(since \(90^\circ - 15^\circ = 75^\circ\))} \]

### Final Answer
Thus, one of the acute angles of triangle ABC is:
\[ \boxed{75^\circ} \]