Question

The angle of elevation of the top of a building from a point A, on the ground, is 30\(^{\circ}\). On moving a distance of 24 m towards its base to the point B, the angle of elevation changes to 60\(^{\circ}\). Find the height of the building and distance of point A from the base of the building. (Take \(\sqrt{3}\) = 1.73)

Hint:

For nested elevation problems where the angle changes from \( 30^{\circ} \) to \( 60^{\circ} \) over a distance \( d \):
The height \( h \) can be directly calculated using the shortcut:
\[ h = \frac{d\sqrt{3}}{2} \]
Here, \( h = \frac{24\sqrt{3}}{2} = 12\sqrt{3}\text{ m} \). This is a helpful mental check to confirm your calculations.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Heights and Distances (Trigonometry Application).
We have a vertical building. From a point \( A \) on the ground, the angle of elevation is \( 30^{\circ} \).
After walking 24 m towards the base to point \( B \), the angle of elevation increases to \( 60^{\circ} \).
We need to find the height of the building and the total distance of point \( A \) from the base.

Step 2: Key Formula or Approach:
- Let the height of the building be \( h \) meters.
- Let \( C \) represent the base of the building, so \( BC = x \) meters.
- Apply the tangent trigonometric ratio in the two right-angled triangles \( \Delta DBC \) and \( \Delta DAC \).

Step 3: Detailed Explanation:
1. Let \( h \) be the height of the building \( CD \).
Let \( x \) be the distance \( BC \) from the second observer point to the base of the building.
The total distance of point \( A \) from the base is \( AC = x + 24 \).
2. In right-angled triangle \( \Delta DBC \) (with angle \( 60^{\circ} \) at \( B \)):
\[ \tan 60^{\circ} = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CD}{BC} \]
\[ \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}} \quad \text{(Equation 1)} \]
3. In right-angled triangle \( \Delta DAC \) (with angle \( 30^{\circ} \) at \( A \)):
\[ \tan 30^{\circ} = \frac{CD}{AC} = \frac{h}{x + 24} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 24} \implies x + 24 = h\sqrt{3} \quad \text{(Equation 2)} \]
4. Substitute \( x = \frac{h}{\sqrt{3}} \) from Equation 1 into Equation 2:
\[ \frac{h}{\sqrt{3}} + 24 = h\sqrt{3} \]
5. Rearrange terms to solve for \( h \):
\[ 24 = h\sqrt{3} - \frac{h}{\sqrt{3}} \]
\[ 24 = h \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \]
\[ 24 = h \left(\frac{3 - 1}{\sqrt{3}}\right) \]
\[ 24 = \frac{2h}{\sqrt{3}} \]
\[ 2h = 24\sqrt{3} \implies h = 12\sqrt{3}\text{ m} \]
6. Calculate the numerical height using \( \sqrt{3} = 1.73 \):
\[ h = 12 \times 1.73 = 20.76\text{ m} \]
7. Find the distance of point \( A \) from the base (\( AC \)):
First, find \( x \):
\[ x = \frac{h}{\sqrt{3}} = \frac{12\sqrt{3}}{\sqrt{3}} = 12\text{ m} \]
The distance from point \( A \) to the base is:
\[ AC = x + 24 = 12 + 24 = 36\text{ m} \]

Step 4: Final Answer:
The height of the building is \(20.76\text{ m}\) and the distance of point A from the base is \(36\text{ m}\).