In a right angled triangle base BC=`15` cm and `sinB=(4)/(5)` , then what is the length of hypotenuse AB ?

Options

A. `25` cm
B. `20` cm
C. `5` cm
D. `4` cm

Answer

To solve the problem, we need to find the length of the hypotenuse \( AB \) in a right-angled triangle where the base \( BC = 15 \) cm and \( \sin B = \frac{4}{5} \).

### Step-by-Step Solution:

1. **Understand the sine ratio**:
The sine of angle \( B \) is defined as the ratio of the length of the opposite side (perpendicular) to the length of the hypotenuse. Therefore, we can express this as:
\[ \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{AB} \]
Given that \( \sin B = \frac{4}{5} \), we can set:
\[ AC = 4x \quad \text{and} \quad AB = 5x \]
where \( x \) is a common multiplier.

**Hint**: Remember that sine relates the opposite side to the hypotenuse.

2. **Identify the base and the sides**:
We know the base \( BC = 15 \) cm. In our triangle, we have:
\[ BC = 15 \text{ cm} \]
We can use the Pythagorean theorem to relate the sides of the triangle.

3. **Apply the Pythagorean theorem**:
According to the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]
Substituting the expressions for \( AC \) and \( AB \):
\[ (5x)^2 = (4x)^2 + (15)^2 \]

4. **Simplify the equation**:
Expanding both sides gives:
\[ 25x^2 = 16x^2 + 225 \]
Rearranging the equation:
\[ 25x^2 - 16x^2 = 225 \]
\[ 9x^2 = 225 \]

5. **Solve for \( x^2 \)**:
Dividing both sides by 9:
\[ x^2 = \frac{225}{9} = 25 \]
Taking the square root:
\[ x = 5 \]

6. **Find the hypotenuse \( AB \)**:
Now, substitute \( x \) back into the expression for the hypotenuse:
\[ AB = 5x = 5 \times 5 = 25 \text{ cm} \]

### Final Answer:
The length of the hypotenuse \( AB \) is \( 25 \) cm.

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