The angular elevation of tower CD at a point A due south of it is `60^(@)` and at a point B due west of A, the elevation is `30^(@)`. If AB = 5 km, then the height of the tower is (where, C, B and A are on the same ground level)

Options

A. `2sqrt3km`
B. `2sqrt6km`
C. `(5sqrt3)/(2)km`
D. `(5sqrt3)/(2sqrt2)km`

Answer

To solve the problem, we will use trigonometric relationships in right triangles formed by the tower and the points A and B.

### Step-by-step Solution:

1. **Understanding the Setup**:
- Let the height of the tower CD be \( h \).
- Point A is due south of the tower, and the angular elevation from A to the top of the tower is \( 60^\circ \).
- Point B is due west of A, and the angular elevation from B to the top of the tower is \( 30^\circ \).
- The distance \( AB = 5 \) km.

2. **Setting Up the Triangles**:
- From point A, we can form a right triangle \( CDA \) where:
- The angle \( \angle CAD = 60^\circ \)
- The opposite side (height of the tower) is \( h \)
- The adjacent side (distance from A to the base of the tower) is \( x \).
- From point B, we can form another right triangle \( CDB \) where:
- The angle \( \angle CBD = 30^\circ \)
- The opposite side (height of the tower) is \( h \)
- The adjacent side (distance from B to the base of the tower) is \( y \).

3. **Using Trigonometric Ratios**:
- For triangle \( CDA \):
\[ \tan(60^\circ) = \frac{h}{x} \implies h = x \cdot \tan(60^\circ) = x \cdot \sqrt{3} \]
- For triangle \( CDB \):
\[ \tan(30^\circ) = \frac{h}{y} \implies h = y \cdot \tan(30^\circ) = y \cdot \frac{1}{\sqrt{3}} \]

4. **Relating the Distances**:
- Since \( B \) is due west of \( A \) and \( AB = 5 \) km, we have:
\[ y = x + 5 \]

5. **Substituting and Solving**:
- From the equations derived from the triangles:
\[ x \cdot \sqrt{3} = (x + 5) \cdot \frac{1}{\sqrt{3}} \]
- Cross-multiplying gives:
\[ x \cdot 3 = x + 5 \]
- Rearranging:
\[ 3x - x = 5 \implies 2x = 5 \implies x = \frac{5}{2} \text{ km} \]

6. **Finding the Height**:
- Now substituting \( x \) back to find \( h \):
\[ h = x \cdot \sqrt{3} = \frac{5}{2} \cdot \sqrt{3} = \frac{5\sqrt{3}}{2} \text{ km} \]

### Final Answer:
The height of the tower CD is \( \frac{5\sqrt{3}}{2} \) km.