The height of a house subtends a right angle at the opposite street light. The angle of elevation of light from the base of the house is `60^(@)`. If the width of the road be 6 meters, then the height of the house is

Options

A. `8sqrt(3)` m
B. 8 m
C. 6 m
D. `6sqrt(3)`m

Answer

To solve the problem step by step, we will use trigonometric ratios and the information provided in the question.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a house and a street light. The height of the house subtends a right angle at the street light. The angle of elevation from the base of the house to the top of the street light is \(60^\circ\). The width of the road is 6 meters.

2. **Define Variables**:
- Let \(h_1\) be the height of the house.
- Let \(h_2\) be the height of the street light.

3. **Use the Angle of Elevation**:
- From the base of the house, the angle of elevation to the street light is \(60^\circ\).
- Using the tangent function:
\[ \tan(60^\circ) = \frac{h_2}{6} \]
- We know that \(\tan(60^\circ) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{h_2}{6} \]

4. **Solve for \(h_2\)**:
- Rearranging the equation gives:
\[ h_2 = 6 \cdot \sqrt{3} \]

5. **Consider the Right Triangle**:
- The height of the house \(h_1\) can be expressed in terms of \(h_2\) and the height above the house:
\[ h_1 - h_2 = \text{height of the house above the street light} \]

6. **Use the Right Triangle with Angle \(30^\circ\)**:
- The angle of elevation from the top of the house to the street light is \(30^\circ\) (since \(90^\circ - 60^\circ = 30^\circ\)).
- Using the tangent function again:
\[ \tan(30^\circ) = \frac{6}{h_1 - h_2} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{6}{h_1 - h_2} \]

7. **Solve for \(h_1 - h_2\)**:
- Rearranging gives:
\[ h_1 - h_2 = 6 \sqrt{3} \]

8. **Substitute \(h_2\) into the Equation**:
- We already found \(h_2 = 6 \sqrt{3}\), so substituting gives:
\[ h_1 - 6\sqrt{3} = 6\sqrt{3} \]

9. **Solve for \(h_1\)**:
- Adding \(6\sqrt{3}\) to both sides gives:
\[ h_1 = 6\sqrt{3} + 6\sqrt{3} = 12\sqrt{3} \]

10. **Final Result**:
- Therefore, the height of the house \(h_1\) is:
\[ h_1 = 12\sqrt{3} \text{ meters} \]