The angle of elevation of a tower at a level ground is `30^(@).` The angle of elevation becomes `theta` when moved 10 m towards the tower. If the height of tower is `5 sqrt3` m, then what is the value of`theta` ?

Options

A. `45 ^(@)`
B. `60^(@)`
C. `75^(@)`
D. None

Answer

To solve the problem step by step, we will use trigonometric principles related to angles of elevation and right triangles.

### Step 1: Understand the problem and draw a diagram
- We have a tower of height \( h = 5\sqrt{3} \) meters.
- From a point \( A \) on the ground, the angle of elevation to the top of the tower is \( 30^\circ \).
- When we move 10 meters closer to the tower to point \( B \), the angle of elevation changes to \( \theta \).

### Step 2: Set up the triangles
- Let \( x \) be the horizontal distance from point \( A \) to the base of the tower.
- Therefore, the distance from point \( B \) to the base of the tower is \( x - 10 \).

### Step 3: Use the tangent function for triangle \( ABC \)
- In triangle \( ABC \) (where \( C \) is the top of the tower), we can write:
\[ \tan(30^\circ) = \frac{h}{x} \]
Substituting the height of the tower:
\[ \tan(30^\circ) = \frac{5\sqrt{3}}{x} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{5\sqrt{3}}{x} \]

### Step 4: Solve for \( x \)
- Cross-multiplying gives:
\[ x = 5\sqrt{3} \cdot \sqrt{3} = 5 \cdot 3 = 15 \]

### Step 5: Use the tangent function for triangle \( ABD \)
- Now, for triangle \( ABD \):
\[ \tan(\theta) = \frac{h}{x - 10} \]
Substituting the known values:
\[ \tan(\theta) = \frac{5\sqrt{3}}{15 - 10} = \frac{5\sqrt{3}}{5} = \sqrt{3} \]

### Step 6: Find \( \theta \)
- Since \( \tan(\theta) = \sqrt{3} \), we find:
\[ \theta = 60^\circ \]

### Final Answer
The value of \( \theta \) is \( 60^\circ \).

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