A man standing 30ms South of a tower of height h walks 60 m to the East and finds the angle of elevation of the top of the tower to `30^(@)`. Find h.

Answer

To solve the problem step by step, we will analyze the situation and apply trigonometric principles.

### Step 1: Understand the Problem
A man is standing 30 meters south of a tower and walks 60 meters east. We need to find the height of the tower (h) given that the angle of elevation from his new position to the top of the tower is 30 degrees.

### Step 2: Set Up the Diagram
1. Let the tower be at point A.
2. The man starts at point B, which is 30 meters south of A.
3. After walking 60 meters east, he reaches point C.
4. The height of the tower is represented as OA (h), and the distance from the base of the tower to point C is AC.

### Step 3: Calculate the Distance AC
Using the Pythagorean theorem, we can find the distance AC:

- The distance AB (south) = 30 m
- The distance BC (east) = 60 m

Using the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
\[ AC^2 = 30^2 + 60^2 \]
\[ AC^2 = 900 + 3600 \]
\[ AC^2 = 4500 \]
\[ AC = \sqrt{4500} = 30\sqrt{5} \text{ m} \]

### Step 4: Use the Angle of Elevation
From point C, the angle of elevation to the top of the tower (point A) is given as 30 degrees. We can use the tangent function to relate the height of the tower (h) to the distance AC:

Using the tangent of the angle:
\[ \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{AC} \]

Substituting the known values:
\[ \tan(30^\circ) = \frac{h}{30\sqrt{5}} \]

### Step 5: Solve for h
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

So we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{30\sqrt{5}} \]

Cross-multiplying gives:
\[ h = 30\sqrt{5} \cdot \frac{1}{\sqrt{3}} \]
\[ h = \frac{30\sqrt{5}}{\sqrt{3}} \]

### Step 6: Simplify h
To simplify:
\[ h = 30 \cdot \frac{\sqrt{5}}{\sqrt{3}} \]
\[ h = 30 \cdot \sqrt{\frac{5}{3}} \]
\[ h = 30 \cdot \sqrt{5} \cdot \frac{\sqrt{3}}{3} \]
\[ h = 10\sqrt{15} \text{ m} \]

### Final Answer
The height of the tower is:
\[ h = 10\sqrt{15} \text{ meters} \]

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