The angular elevations of a tower CD at a place A due south of it is `60^(@)` , and at a place B due west of A, the elevation is `30^(@)` . If AB=300m, what is the height, in meters, of the tower?

Options

A. 30
B. `30sqrt(6)`
C. `30sqrt(10)`
D. `30sqrt(30)`

Answer

To solve the problem, we will follow these steps:

1. **Understanding the Setup**:
- We have a tower CD.
- Point A is due south of the tower CD, and the angle of elevation from A to the top of the tower is \(60^\circ\).
- Point B is due west of A, and the angle of elevation from B to the top of the tower is \(30^\circ\).
- The distance \(AB = 300\) m.

2. **Identifying the Triangles**:
- We will analyze two right triangles:
- Triangle \(ADC\) (where \(D\) is the base of the tower and \(C\) is the top of the tower).
- Triangle \(BCD\) (where \(B\) is the point due west of \(A\)).

3. **Setting Up the Heights and Bases**:
- Let the height of the tower \(CD = h\).
- In triangle \(ADC\):
- The angle of elevation from \(A\) is \(60^\circ\).
- We can use the tangent function:
\[ \tan(60^\circ) = \frac{h}{AD} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h}{AD} \implies AD = \frac{h}{\sqrt{3}} \]

- In triangle \(BCD\):
- The angle of elevation from \(B\) is \(30^\circ\).
- Again using the tangent function:
\[ \tan(30^\circ) = \frac{h}{BD} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{BD} \implies BD = h\sqrt{3} \]

4. **Using the Pythagorean Theorem**:
- In triangle \(ABD\), we can apply the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
Substituting \(AB = 300\) m, \(AD = \frac{h}{\sqrt{3}}\), and \(BD = h\sqrt{3}\):
\[ 300^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + (h\sqrt{3})^2 \]
Simplifying this gives:
\[ 90000 = \frac{h^2}{3} + 3h^2 \]
\[ 90000 = \frac{h^2 + 9h^2}{3} = \frac{10h^2}{3} \]
Multiplying both sides by 3:
\[ 270000 = 10h^2 \implies h^2 = 27000 \implies h = \sqrt{27000} \]

5. **Calculating the Height**:
- Simplifying \(\sqrt{27000}\):
\[ \sqrt{27000} = \sqrt{27 \times 1000} = \sqrt{27} \times \sqrt{1000} = 3\sqrt{3} \times 10\sqrt{10} = 30\sqrt{30} \]

Thus, the height of the tower \(CD\) is \(30\sqrt{30}\) meters.