The angle of elevation of an object on a hill from a point on the ground is `30^@`. After walking 120 metres the elevation of the object is `60^@`. The height of the hill is

Options

A. 120 m
B. `60sqrt3` m
C. `120sqrt3` m
D. 60 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. **Understanding the Problem:**
- We have a hill (let's denote it as AB) and two points on the ground (let's denote them as C and D).
- From point C, the angle of elevation to the top of the hill (point A) is \(30^\circ\).
- After walking 120 meters towards the hill to point D, the angle of elevation to point A becomes \(60^\circ\).

2. **Setting Up the Diagram:**
- Let the height of the hill (AB) be \(h\).
- Let the horizontal distance from point C to the base of the hill (point B) be \(x\).
- Therefore, the distance from point D to point B is \(x - 120\) meters.

3. **Using Triangle ACB (Angle \(60^\circ\)):**
- In triangle ACB, we can use the tangent function:
\[ \tan(60^\circ) = \frac{h}{x} \]
- We know that \(\tan(60^\circ) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x \quad \text{(Equation 1)} \]

4. **Using Triangle ADB (Angle \(30^\circ\)):**
- In triangle ADB, we can use the tangent function again:
\[ \tan(30^\circ) = \frac{h}{x - 120} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x - 120} \implies h = \frac{1}{\sqrt{3}}(x - 120) \quad \text{(Equation 2)} \]

5. **Equating the Two Expressions for \(h\):**
- From Equation 1 and Equation 2, we have:
\[ \sqrt{3}x = \frac{1}{\sqrt{3}}(x - 120) \]
- Multiplying both sides by \(\sqrt{3}\) to eliminate the fraction:
\[ 3x = x - 120 \]
- Rearranging gives:
\[ 3x - x = -120 \implies 2x = 120 \implies x = 60 \]

6. **Finding the Height \(h\):**
- Now substitute \(x = 60\) back into Equation 1:
\[ h = \sqrt{3} \cdot 60 = 60\sqrt{3} \]

### Final Answer:
The height of the hill is \(60\sqrt{3}\) meters.