If the hypotenuse of a right-angled triangle is four times the length of the perpendicular drawn from the opposite vertex to it, then the difference of the two acute angles will be `60^0` (b) `15^0` (c) `75^0` (d) `30^0`

Options

A. `60^(@)`
B. `15^(@)`
C. `75^(@)`
D. `30^(@)`

Explanation

To solve the problem step by step, we start by analyzing the given information about the right-angled triangle.

### Step 1: Define the Variables
Let:
- \( AB \) be the length of the hypotenuse.
- \( CD \) be the length of the perpendicular drawn from the opposite vertex to the hypotenuse.
- According to the problem, \( AB = 4 \times CD \).

### Step 2: Express in Terms of a Variable
Let \( CD = P \). Then, we can express the hypotenuse as:
\[ AB = 4P \]

### Step 3: Calculate the Area of the Triangle
The area \( A \) of a right-angled triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, the base can be taken as \( AB \) and the height as \( CD \):
\[ A = \frac{1}{2} \times AB \times CD = \frac{1}{2} \times (4P) \times P = 2P^2 \]

### Step 4: Use the Pythagorean Theorem
In a right triangle, the relationship between the sides is given by:
\[ A^2 + B^2 = C^2 \]
Where \( C \) is the hypotenuse. Here, we can denote the other two sides as \( A \) and \( B \). Therefore:
\[ A^2 + B^2 = (4P)^2 = 16P^2 \]

### Step 5: Express \( A^2 + B^2 \) in Terms of \( P \)
We also know that:
\[ A^2 + B^2 = 16P^2 \]
We can express \( A \) and \( B \) in terms of \( P \) using the area:
\[ A \times B = 2P^2 \]

### Step 6: Set Up the Equations
We have two equations:
1. \( A^2 + B^2 = 16P^2 \)
2. \( A \times B = 2P^2 \)

### Step 7: Use the Identity
Using the identity \( (A - B)^2 = A^2 + B^2 - 2AB \):
\[ (A - B)^2 = 16P^2 - 4P^2 = 12P^2 \]
Thus,
\[ A - B = \sqrt{12}P = 2\sqrt{3}P \]

### Step 8: Use the Other Identity
Using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \):
\[ (A + B)^2 = 16P^2 + 4P^2 = 20P^2 \]
Thus,
\[ A + B = \sqrt{20}P = 2\sqrt{5}P \]

### Step 9: Find the Difference of Angles
Using the formulas for the angles in terms of \( A \) and \( B \):
\[ \tan\left(\frac{A - B}{2}\right) = \frac{A - B}{A + B} \cdot \cot\left(\frac{C}{2}\right) \]
We know that \( \cot\left(\frac{C}{2}\right) = \frac{1}{\sqrt{3}} \) (since \( C = 90^\circ \)).

Substituting the values:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{2\sqrt{3}P}{2\sqrt{5}P} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{5}} \]

### Step 10: Calculate the Angle
Taking the inverse tangent:
\[ \frac{A - B}{2} = 30^\circ \implies A - B = 60^\circ \]

### Conclusion
The difference of the two acute angles is:
\[ \boxed{60^\circ} \]