The angle of elevation of a tower at a level ground is `30^(@)`. The angle of elevation becomes `theta` when 10 m moved towards the tower. If the height of tower is `5sqrt(3)` m, then what is `theta` equal to ?

Options

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

Explanation

To solve the problem, we will use the concept of right triangles and trigonometric ratios. Let's break down the solution step by step.

### Step 1: Understand the situation
We have a tower of height \( h = 5\sqrt{3} \) meters. The angle of elevation from point A (at a distance from 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 triangle
Let the distance from point A to the base of the tower be \( d \). From point A, using the angle of elevation \( 30^\circ \), we can write:

\[ \tan(30^\circ) = \frac{\text{height of tower}}{\text{distance from tower}} = \frac{5\sqrt{3}}{d} \]

### Step 3: Solve for \( d \)
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Thus, we can set up the equation:

\[ \frac{1}{\sqrt{3}} = \frac{5\sqrt{3}}{d} \]

Cross-multiplying gives:

\[ d = 5\sqrt{3} \cdot \sqrt{3} = 5 \cdot 3 = 15 \text{ meters} \]

### Step 4: Find the new distance from point B
When we move 10 meters closer to the tower, the new distance \( d' \) from point B to the tower is:

\[ d' = d - 10 = 15 - 10 = 5 \text{ meters} \]

### Step 5: Set up the equation for angle \( \theta \)
Now, we can find \( \theta \) using the new distance \( d' \):

\[ \tan(\theta) = \frac{\text{height of tower}}{\text{new distance}} = \frac{5\sqrt{3}}{5} \]

### Step 6: Simplify the equation
This simplifies to:

\[ \tan(\theta) = \sqrt{3} \]

### Step 7: Find \( \theta \)
The angle whose tangent is \( \sqrt{3} \) is:

\[ \theta = 60^\circ \]

### Final Answer
Thus, the value of \( \theta \) is \( 60^\circ \).

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