Find the angle of elevation of a point which is at a distance of 30 m from the base of a tower `10sqrt(3)` m high.

Answer

To find the angle of elevation of a point that is at a distance of 30 m from the base of a tower that is \(10\sqrt{3}\) m high, we can follow these steps:

### Step 1: Understand the problem
We have a tower of height \(10\sqrt{3}\) m and a point on the ground that is 30 m away from the base of the tower. We need to find the angle of elevation from this point to the top of the tower.

### Step 2: Draw a diagram
Let's visualize the situation:
- Draw a vertical line representing the tower (height = \(10\sqrt{3}\) m).
- Draw a horizontal line from the base of the tower to the point on the ground (distance = 30 m).
- The angle of elevation (\(\theta\)) is formed between the line of sight to the top of the tower and the horizontal line.

### Step 3: Identify the right triangle
We can form a right triangle where:
- The height of the tower is the opposite side (perpendicular) = \(10\sqrt{3}\) m.
- The distance from the point to the base of the tower is the adjacent side (base) = 30 m.

### Step 4: Use the tangent function
The tangent of the angle of elevation (\(\theta\)) is given by the ratio of the opposite side to the adjacent side:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{10\sqrt{3}}{30} \]

### Step 5: Simplify the expression
Now, simplify the fraction:
\[ \tan(\theta) = \frac{10\sqrt{3}}{30} = \frac{\sqrt{3}}{3} \]

### Step 6: Find the angle
The value of \(\tan(\theta) = \frac{\sqrt{3}}{3}\) corresponds to an angle of:
\[ \theta = 30^\circ \]

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

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