What is the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height `10sqrt(3)`m?

Answer

To find the angle of elevation of the top of a tower from a point on the ground, we can follow these steps:

### Step 1: Understand the problem
We have a tower of height \( h = 10\sqrt{3} \) meters and a point on the ground that is \( d = 30 \) 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: Set up the right triangle
We can visualize this situation as a right triangle where:
- The height of the tower is the opposite side (perpendicular) to the angle \( \theta \).
- The distance from the point on the ground to the foot of the tower is the adjacent side (base) of the triangle.

### Step 3: Use the tangent function
The tangent of the angle \( \theta \) is given by the ratio of the opposite side to the adjacent side:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d} \]
Substituting the values we have:
\[ \tan(\theta) = \frac{10\sqrt{3}}{30} \]

### Step 4: Simplify the expression
Now, we simplify the right-hand side:
\[ \tan(\theta) = \frac{10\sqrt{3}}{30} = \frac{\sqrt{3}}{3} \]

### Step 5: Find the angle \( \theta \)
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(60^\circ) = \sqrt{3} \]
Thus, we can find \( \theta \) using the known values:
\[ \tan(60^\circ) = \sqrt{3} \quad \text{and} \quad \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Since \( \tan(30^\circ) = \frac{\sqrt{3}}{3} \), we conclude:
\[ \theta = 30^\circ \]

### Final Answer
The angle of elevation of the top of the tower is \( 30^\circ \).

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