The angle of elevation of the top of a tower at a point X is 30°. On walking 30 m towards the tower if the angle of elevation becomes 60° at point Y, find the height of the tower?

Options

A. 30 m
B. 29 m
C. 28 m
D. 26 m

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We have a tower (let's denote the height of the tower as AB) and two points, X and Y. The angle of elevation from point X to the top of the tower is 30°, and after walking 30 meters towards the tower to point Y, the angle of elevation becomes 60°.

### Step 2: Set Up the Diagram
Draw a vertical line representing the tower (AB) and horizontal lines from points X and Y to the base of the tower (A). The distance from point X to point A is denoted as XA, and the distance from point Y to point A is XA - 30 meters.

### Step 3: Use Trigonometric Ratios
From point X:
- The angle of elevation is 30°.
- Using the tangent function, we have:
\[ \tan(30°) = \frac{AB}{XA} \]
We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\), so:
\[ \frac{1}{\sqrt{3}} = \frac{AB}{XA} \]
Rearranging gives us:
\[ AB = \frac{XA}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 4: Apply the Second Angle of Elevation
From point Y:
- The angle of elevation is 60°.
- Again using the tangent function:
\[ \tan(60°) = \frac{AB}{XA - 30} \]
We know that \(\tan(60°) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{AB}{XA - 30} \]
Rearranging gives us:
\[ AB = \sqrt{3}(XA - 30) \quad \text{(Equation 2)} \]

### Step 5: Set Equations Equal
Now we have two expressions for AB:
1. From Equation 1: \(AB = \frac{XA}{\sqrt{3}}\)
2. From Equation 2: \(AB = \sqrt{3}(XA - 30)\)

Setting them equal to each other:
\[ \frac{XA}{\sqrt{3}} = \sqrt{3}(XA - 30) \]

### Step 6: Solve for XA
Cross-multiplying gives:
\[ XA = 3(XA - 30) \]
Expanding this:
\[ XA = 3XA - 90 \]
Rearranging:
\[ 90 = 3XA - XA \]
\[ 90 = 2XA \]
\[ XA = 45 \text{ meters} \]

### Step 7: Find the Height of the Tower (AB)
Now substituting \(XA\) back into Equation 1 to find AB:
\[ AB = \frac{45}{\sqrt{3}} = 15\sqrt{3} \]
Calculating \(15\sqrt{3}\) gives approximately:
\[ AB \approx 15 \times 1.732 \approx 25.98 \text{ meters} \]
So, the height of the tower is approximately 26 meters.

### Final Answer
The height of the tower is approximately **26 meters**.

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