A tower stands at the centre of a circular park . A and B are two points on the boundary of the park such that AB(=a) subtends an angle of `60^@` at the foot of the tower , and the angle of elevation of the top of the tower from A or B is `30^@`. The height of the tower is

Options

A. `a//sqrt(3)`
B. `asqrt(3)`
C. `2a//sqrt(3)`
D. `2asqrt(3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a circular park with a tower at the center. Points A and B are on the boundary of the park, and the line segment AB subtends an angle of \(60^\circ\) at the foot of the tower (let's call it point P). The angle of elevation from points A and B to the top of the tower (point Q) is \(30^\circ\).

### Step 2: Label the Diagram
- Let the length of segment AB be \(a\).
- Since AB subtends an angle of \(60^\circ\) at P, we can denote the midpoint of AB as M. Thus, AM = MB = \( \frac{a}{2} \).
- The angles \( \angle APM \) and \( \angle BPM \) are both \(30^\circ\) because the total angle \( \angle APB \) is \(60^\circ\).

### Step 3: Use Trigonometry in Triangle AMP
In triangle AMP, we can use the sine function:
\[ \sin(30^\circ) = \frac{AM}{AP} \]
Here, \(AM = \frac{a}{2}\) and \(AP\) is the hypotenuse. Thus, we have:
\[ \sin(30^\circ) = \frac{\frac{a}{2}}{AP} \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we can set up the equation:
\[ \frac{1}{2} = \frac{\frac{a}{2}}{AP} \]
Cross-multiplying gives:
\[ AP = a \]

### Step 4: Use Trigonometry in Triangle PAQ
Now, in triangle PAQ, we can use the tangent function:
\[ \tan(30^\circ) = \frac{PQ}{AP} \]
Here, \(PQ\) is the height of the tower, and \(AP = a\). Thus, we have:
\[ \tan(30^\circ) = \frac{PQ}{a} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can set up the equation:
\[ \frac{1}{\sqrt{3}} = \frac{PQ}{a} \]
Cross-multiplying gives:
\[ PQ = \frac{a}{\sqrt{3}} \]

### Step 5: Conclusion
The height of the tower \(PQ\) is given by:
\[ \boxed{\frac{a}{\sqrt{3}}} \]