Question

The angles of elevation and depression of the top and bottom of a lighthouse from the top of a 60 m high building are 30º and 60º respectively. Then the difference between the heights of the lighthouse and building is

• A. 20 m
• B. 80 m
• C. 60 m
• D. 40 m

Answer 1

The Correct Option is A

Given:

- Height of the  building = \( 60 \) m 
- Angle of elevation to the top of the lighthouse = \( 30^\circ \) 
- Angle of depression to the base of the lighthouse = \( 60^\circ \)

Step 1: Define Variables

Let the height of the lighthouse be \( H \) meters, and let the horizontal distance between the lighthouse and the building be \( d \) meters.

Step 2: Use the Tangent Function

For the angle of elevation to the top of the lighthouse: \[ \tan 30^\circ = \frac{H - 60}{d} \] \[ \frac{1}{\sqrt{3}} = \frac{H - 60}{d} \] \[ H - 60 = \frac{d}{\sqrt{3}} \] For the angle of depression to the base of the lighthouse: \[ \tan 60^\circ = \frac{60}{d} \] \[ \sqrt{3} = \frac{60}{d} \] \[ d = \frac{60}{\sqrt{3}} = 20\sqrt{3} \]

Step 3: Solve for \( H \)

\[ H - 60 = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \] \[ H = 80 \]

Step 4: Compute the Difference in Heights

\[ H - 60 = 80 - 60 = 20 \]

Final Answer: 20 m

Answer 2

The problem involves using trigonometry to find the difference in height between a lighthouse and a 60 m high building. We analyze the angles of elevation and depression observed from the top of the building.

1. Let the height of the lighthouse be \( h \) meters. The angles given are from the top of the building.
2. The angle of elevation of the top of the lighthouse is 30º, and the angle of depression to the bottom is 60º.
3. Let the distance from the base of the building to the base of the lighthouse be \( d \) meters.
4. Using trigonometry, when the angle of elevation is 30º: 
   \(\tan(30º) = \frac{h-60}{d}\)
   \(\frac{1}{\sqrt{3}} = \frac{h-60}{d}\)
   so, \(h-60 = \frac{d}{\sqrt{3}}\)
5. For the angle of depression 60º: 
   \(\tan(60º) = \frac{60}{d}\)
   \(\sqrt{3} = \frac{60}{d}\) 
   Therefore, \(d = \frac{60}{\sqrt{3}}\)
6. Equating \(d\) from both equations gives: 
   \(\frac{60}{\sqrt{3}} = d = \sqrt{3} \times (h-60)\)
7. Substitute \(d = \frac{60}{\sqrt{3}}\) into equation for \(h\):
   \(\frac{60}{\sqrt{3}} = \sqrt{3} \times (h-60)\)
   Multiplying throughout by \(\sqrt{3}\), we get \(60 = 3(h-60)\)
8. Solving for \(h\), we have:
   \(60 = 3h - 180\) 
   Therefore, \(3h = 240\) 
   \(h = 80\)
9. The height difference between the lighthouse and building is \(80 - 60 = 20\) meters.

Thus, the difference between the heights of the lighthouse and the building is 20 m.