The angle of depression of the foot of a building from the top of a tower `32sqrt3` m high is `60^@` How far away from the 6 building is the tower?

Options

A. 32 m
B. 16 m
C. `16sqrt3` m
D. `32sqrt3` m

Answer

To solve the problem step by step, we will use the properties of right triangles and trigonometric ratios.

### Step 1: Understand the Problem
We have a tower of height \( h = 32\sqrt{3} \) m. The angle of depression from the top of the tower to the foot of the building is \( 60^\circ \). We need to find the horizontal distance \( x \) from the base of the tower to the base of the building.

### Step 2: Draw a Diagram
Draw a vertical line representing the tower and label its height as \( 32\sqrt{3} \) m. Draw a horizontal line from the top of the tower to the foot of the building. The angle of depression is \( 60^\circ \), which means the angle between the horizontal line and the line of sight to the foot of the building is also \( 60^\circ \).

### Step 3: Identify the Right Triangle
From the top of the tower to the foot of the building, we can form a right triangle:
- The height of the tower is the opposite side to the angle of depression.
- The distance from the tower to the building is the adjacent side.

### Step 4: Use the Tangent Function
Using the tangent function, we know that:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, \( \theta = 60^\circ \), the opposite side is \( h = 32\sqrt{3} \), and the adjacent side is \( x \):
\[ \tan(60^\circ) = \frac{32\sqrt{3}}{x} \]

### Step 5: Substitute the Value of \( \tan(60^\circ) \)
We know that \( \tan(60^\circ) = \sqrt{3} \). Substituting this into the equation gives:
\[ \sqrt{3} = \frac{32\sqrt{3}}{x} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ \sqrt{3} \cdot x = 32\sqrt{3} \]
Now, divide both sides by \( \sqrt{3} \):
\[ x = 32 \]

### Step 7: Conclusion
The distance from the tower to the foot of the building is \( 32 \) meters.

### Final Answer
The tower is \( 32 \) meters away from the building.

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