The angles of elevation of the top of a tower at the top and the foot of a pole of height 10 m are `30^@and 60^@` respectively. The height of the tower is

Options

A. 10 m
B. 15 m
C. 20 m
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a pole of height 10 m and a tower. The angles of elevation to the top of the tower from the top of the pole and from the ground are given as \(30^\circ\) and \(60^\circ\) respectively. We need to find the height of the tower.

### Step 2: Set Up the Diagram
Let's denote:
- The height of the pole as \(AB = 10\) m.
- The height of the tower as \(DC = h + 10\) m (where \(h\) is the height of the tower above the pole).
- The distance from the foot of the pole to the foot of the tower as \(x\).

### Step 3: Use Trigonometric Ratios
1. **From the top of the pole (point A)**, we have:
\[ \tan(30^\circ) = \frac{DC - AB}{x} = \frac{h}{x} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can write:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \implies x = h\sqrt{3} \quad \text{(Equation 1)} \]

2. **From the ground (point B)**, we have:
\[ \tan(60^\circ) = \frac{DC}{x} = \frac{h + 10}{x} \]
Since \(\tan(60^\circ) = \sqrt{3}\), we can write:
\[ \sqrt{3} = \frac{h + 10}{x} \implies x = \frac{h + 10}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Expressions for \(x\)
From Equation 1 and Equation 2, we have:
\[ h\sqrt{3} = \frac{h + 10}{\sqrt{3}} \]

### Step 5: Cross-Multiply to Solve for \(h\)
Cross-multiplying gives:
\[ h\sqrt{3} \cdot \sqrt{3} = h + 10 \]
\[ 3h = h + 10 \]

### Step 6: Rearrange the Equation
Rearranging gives:
\[ 3h - h = 10 \implies 2h = 10 \implies h = 5 \]

### Step 7: Calculate the Total Height of the Tower
The total height of the tower \(DC\) is:
\[ DC = h + 10 = 5 + 10 = 15 \text{ m} \]

### Final Answer
The height of the tower is **15 m**.

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