A tower subtends an angle of `30^@` at a point on the same level as the foot of the tower. At a second point h meter above the first, the depression of the foot of the tower is `60^@`. The horizontal distance of the tower from the point is

Options

A. `h/sqrt(3)`
B. `h/(3)`
C. 3h
D. `(3h)/d`

Answer

To solve the problem step by step, we will use trigonometric ratios and the properties of angles of elevation and depression.

### Step 1: Understand the Problem
We have a tower (let's denote its height as AB) and two points (let's denote them as C and D). Point C is on the same level as the foot of the tower (point A), and point D is h meters above point C. At point C, the angle of elevation to the top of the tower (point B) is 30 degrees. At point D, the angle of depression to the foot of the tower (point A) is 60 degrees.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the tower (AB).
2. Mark point A at the base of the tower and point B at the top of the tower.
3. Draw a horizontal line from point A to point C, which is at the same level as A.
4. Mark point D above point C, h meters higher.
5. Draw lines from points C and D to points A and B, indicating the angles of elevation and depression.

### Step 3: Set Up the Trigonometric Relationships
1. From point C, the angle of elevation to point B is 30 degrees. Thus, we can use the tangent function:
\[ \tan(30^\circ) = \frac{AB}{AC} \]
where \( AC \) is the horizontal distance we need to find (let's denote it as \( x \)) and \( AB \) is the height of the tower.

2. From point D, the angle of depression to point A is 60 degrees. The angle of depression is equal to the angle of elevation from point D to point A, which is also 60 degrees. Thus, we can use the tangent function again:
\[ \tan(60^\circ) = \frac{AB + h}{DC} \]
where \( DC \) is the horizontal distance from D to A, which is still \( x \).

### Step 4: Solve the Equations
1. From the first equation:
\[ \tan(30^\circ) = \frac{AB}{x} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{AB}{x} \Rightarrow AB = \frac{x}{\sqrt{3}} \]

2. From the second equation:
\[ \tan(60^\circ) = \frac{AB + h}{x} \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{AB + h}{x} \Rightarrow AB + h = \sqrt{3}x \]

### Step 5: Substitute and Solve for x
Now we can substitute \( AB \) from the first equation into the second equation:
\[ \frac{x}{\sqrt{3}} + h = \sqrt{3}x \]
Multiply through by \( \sqrt{3} \) to eliminate the fraction:
\[ x + h\sqrt{3} = 3x \]
Rearranging gives:
\[ h\sqrt{3} = 3x - x \Rightarrow h\sqrt{3} = 2x \]
Thus:
\[ x = \frac{h\sqrt{3}}{2} \]

### Final Answer
The horizontal distance of the tower from the point is:
\[ x = \frac{h\sqrt{3}}{2} \]