The angle of elevation of the top of a tower of height `100 sqrt(3) m` from a point at a distance of 100 m from the foot of the tower on a horizontal plane is

Options

A. `60^(@)`
B. `90^(@)`
C. `120^(@)`
D. `180^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the situation
We have a tower of height \( h = 100\sqrt{3} \) meters. We are observing the tower from a point that is \( d = 100 \) meters away from the foot of the tower. We need to find the angle of elevation \( \theta \) from this point to the top of the tower.

### Step 2: Draw a diagram
To visualize the problem, we can draw a right triangle where:
- The height of the tower represents the perpendicular side (opposite to the angle \( \theta \)).
- The distance from the foot of the tower to the observation point represents the base (adjacent to the angle \( \theta \)).
- The angle of elevation \( \theta \) is the angle formed between the line of sight to the top of the tower and the horizontal line from the observation point to the foot of the tower.

### Step 3: Use the tangent function
The tangent of the angle \( \theta \) is defined as the ratio of the opposite side (height of the tower) to the adjacent side (distance from the tower):
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \]
Substituting the known values:
\[ \tan(\theta) = \frac{100\sqrt{3}}{100} \]

### Step 4: Simplify the expression
Now, we simplify the right side:
\[ \tan(\theta) = \sqrt{3} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse tangent (arctan) of \( \sqrt{3} \):
\[ \theta = \tan^{-1}(\sqrt{3}) \]

### Step 6: Determine the angle
From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \]
Thus,
\[ \theta = 60^\circ \]

### Final Answer
The angle of elevation of the top of the tower is \( 60^\circ \).

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