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 metre 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. `hcot60^(@)`
B. `(1)/(3)hcot 30^(@)`
C. `(h)/(3)cot60^(@)`
D. `hcot30^(@)`

Answer

To solve the problem, we will use trigonometry concepts related to angles of elevation and depression. Let's break it down step by step.

### Step 1: Understand the Problem
We have a tower that subtends an angle of \(30^\circ\) at a point on the same level as the foot of the tower. At a second point \(h\) meters above the first point, the angle of depression to the foot of the tower is \(60^\circ\). We need to find the horizontal distance from the point to the tower.

### Step 2: Set Up the Diagram
1. Let \(A\) be the foot of the tower.
2. Let \(B\) be the top of the tower.
3. Let \(C\) be the first point on the ground where the angle of elevation to the top of the tower is \(30^\circ\).
4. Let \(D\) be the second point, which is \(h\) meters above point \(C\), where the angle of depression to point \(A\) is \(60^\circ\).

### Step 3: Define Variables
- Let \(x\) be the horizontal distance from point \(C\) to the foot of the tower \(A\).
- Let \(h\) be the height of point \(D\) above point \(C\).

### Step 4: Use Trigonometric Ratios
1. From point \(C\) (at height 0), using the angle of elevation \(30^\circ\):
\[ \tan(30^\circ) = \frac{h}{x} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h \sqrt{3} \]

2. From point \(D\) (at height \(h\)), using the angle of depression \(60^\circ\):
\[ \tan(60^\circ) = \frac{h}{x} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we have:
\[ \sqrt{3} = \frac{h}{x} \]
Rearranging gives:
\[ x = \frac{h}{\sqrt{3}} \]

### Step 5: Equate the Two Expressions for \(x\)
From the two equations we derived for \(x\):
1. \(x = h \sqrt{3}\)
2. \(x = \frac{h}{\sqrt{3}}\)

Setting them equal to each other:
\[ h \sqrt{3} = \frac{h}{\sqrt{3}} \]

### Step 6: Solve for \(x\)
To find the horizontal distance \(x\):
1. From the first equation:
\[ x = h \sqrt{3} \]
2. From the second equation:
\[ x = h \cot(60^\circ) \quad \text{(since } \cot(60^\circ) = \frac{1}{\sqrt{3}} \text{)} \]

Thus, the horizontal distance from the point to the tower is:
\[ x = h \cot(60^\circ) \]

### Final Answer
The horizontal distance of the tower from the point is \(h \cot(60^\circ)\).

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