In `DeltaABC`, a=12 m, `angleB=30^(@)` and `angleC=90^(@)`, then area of `DeltaABC=?`

Options

A. `2npi+ (5pi)/(12)`
B. `2npi-(5pi)/(12)`
C. `npi+(5pi)/(12)`
D. `npi-(5pi)/(12)`

Answer

To find the area of triangle ABC where \( a = 12 \, \text{m} \), \( \angle B = 30^\circ \), and \( \angle C = 90^\circ \), we can use the formula for the area of a triangle based on its sides and angles.

### Step-by-Step Solution:

1. **Identify the angles**:
- Given \( \angle C = 90^\circ \), this is a right triangle.
- The remaining angle \( \angle A \) can be calculated as follows:
\[ \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 30^\circ - 90^\circ = 60^\circ \]

2. **Use the area formula for a triangle**:
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times a \times b \times \sin(C) \]
where \( a \) is the side opposite angle A, \( b \) is the side opposite angle B, and \( C \) is the angle between sides \( a \) and \( b \).

3. **Determine the sides**:
Since \( a = 12 \, \text{m} \) is opposite angle A, we need to find side \( b \) (which is opposite angle B). We can use the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Plugging in the known values:
\[ \frac{12}{\sin(60^\circ)} = \frac{b}{\sin(30^\circ)} \]
Knowing that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ \frac{12}{\frac{\sqrt{3}}{2}} = \frac{b}{\frac{1}{2}} \]
Cross-multiplying gives:
\[ 12 \cdot \frac{1}{2} = b \cdot \frac{\sqrt{3}}{2} \implies 6 = \frac{b \sqrt{3}}{2} \implies b = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

4. **Calculate the area**:
Now we can substitute \( a \), \( b \), and \( C \) into the area formula:
\[ A = \frac{1}{2} \times 12 \times 4\sqrt{3} \times \sin(90^\circ) \]
Since \( \sin(90^\circ) = 1 \):
\[ A = \frac{1}{2} \times 12 \times 4\sqrt{3} \times 1 = 24\sqrt{3} \, \text{m}^2 \]

### Final Answer:
The area of triangle ABC is \( 24\sqrt{3} \, \text{m}^2 \).