If the distance of a point from the base of a building is 15 m and angle of elevation to its top from that point is `60^@` . then what is the height of building?

Options

A. 25 m
B. 26 m
C. 31 m
D. 18 m

Answer

To find the height of the building given the distance from the point to the base and the angle of elevation, we can use trigonometric ratios. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a right triangle formed by the height of the building (H), the distance from the point to the base of the building (15 m), and the line of sight from the point to the top of the building. The angle of elevation from the point to the top of the building is 60 degrees.

### Step 2: Identify the Right Triangle
In the right triangle:
- The height of the building (H) is the opposite side to the angle of elevation.
- The distance from the point to the base of the building (15 m) is the adjacent side.
- The angle of elevation is 60 degrees.

### Step 3: Use the Tangent Function
We can use the tangent function, which is defined as:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
For our triangle:
\[ \tan(60^\circ) = \frac{H}{15} \]

### Step 4: Find the Value of \(\tan(60^\circ)\)
From trigonometric tables or a calculator, we know:
\[ \tan(60^\circ) = \sqrt{3} \]

### Step 5: Set Up the Equation
Substituting the value of \(\tan(60^\circ)\) into the equation gives:
\[ \sqrt{3} = \frac{H}{15} \]

### Step 6: Solve for H
To find H, multiply both sides by 15:
\[ H = 15 \cdot \sqrt{3} \]

### Step 7: Calculate the Height
Now, we can calculate the approximate value of \(\sqrt{3}\):
\[ \sqrt{3} \approx 1.732 \]
So,
\[ H \approx 15 \cdot 1.732 \approx 25.98 \]

### Step 8: Round the Height
Rounding 25.98 to the nearest whole number gives:
\[ H \approx 26 \text{ m} \]

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

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