A radio transmitter antenna of height 100 m stands at the top of a tall building. At a point on the ground, the angle of elevationof the bottom of the antenna is `45^(@)` and that of the the top of the antenna is `60^(@)` . What is the height of the building ?

Options

A. 100 m
B. 50 m
C. `50 (sqrt(3) + 1) m `
D. ` 50 ( sqrt(3) - 1) m `

Answer

To find the height of the building, we can follow these steps:

1. **Understand the Problem**: We have a tall building with an antenna on top. The height of the antenna is 100 m. We need to find the height of the building (let's denote it as \( h \)). We have two angles of elevation from a point on the ground: \( 45^\circ \) for the bottom of the antenna and \( 60^\circ \) for the top of the antenna.

2. **Set Up the Diagram**:
- Let point A be the point on the ground.
- Let point B be the bottom of the antenna (which is at the top of the building).
- Let point C be the top of the antenna.
- The height of the building is \( h \), and the height of the antenna is 100 m, so the total height from point A to point C is \( h + 100 \).

3. **Use Trigonometry for the First Triangle (Angle \( 45^\circ \))**:
- In triangle \( ABE \) (where E is the base of the building), the angle of elevation to point B is \( 45^\circ \).
- Using the tangent function, we have:
\[ \tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{x} \]
- Since \( \tan(45^\circ) = 1 \), we can write:
\[ h = x \quad \text{(Equation 1)} \]

4. **Use Trigonometry for the Second Triangle (Angle \( 60^\circ \))**:
- In triangle \( ACD \) (where D is the top of the antenna), the angle of elevation to point C is \( 60^\circ \).
- Using the tangent function again, we have:
\[ \tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h + 100}{x} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we can write:
\[ \sqrt{3} = \frac{h + 100}{x} \quad \text{(Equation 2)} \]

5. **Substitute Equation 1 into Equation 2**:
- From Equation 1, we know \( x = h \).
- Substitute \( x \) in Equation 2:
\[ \sqrt{3} = \frac{h + 100}{h} \]
- Rearranging gives:
\[ \sqrt{3} h = h + 100 \]

6. **Solve for \( h \)**:
- Rearranging the equation:
\[ \sqrt{3} h - h = 100 \]
\[ h(\sqrt{3} - 1) = 100 \]
\[ h = \frac{100}{\sqrt{3} - 1} \]

7. **Rationalize the Denominator**:
- Multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ h = \frac{100(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{100(\sqrt{3} + 1)}{3 - 1} = \frac{100(\sqrt{3} + 1)}{2} \]
- Simplifying gives:
\[ h = 50(\sqrt{3} + 1) \]

8. **Final Answer**:
- Therefore, the height of the building is \( 50(\sqrt{3} + 1) \) meters.