As observed from the top of a light house, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from `30^(@)` to `45^(@)` . Determine the distance travelled by the ship during the period of observation .

Options

A. 75.3 m
B. 68.59 m
C. 73.2m
D. 74.56m

Answer

To solve the problem step by step, we will use trigonometric concepts related to angles of depression and right triangles.

### Step 1: Understand the Setup
We have a lighthouse that is 100 meters tall. The angle of depression to a ship changes from 30° to 45° as the ship sails towards the lighthouse.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the lighthouse (100 m).
2. From the top of the lighthouse, draw two lines representing the angles of depression: one at 30° and the other at 45°.
3. Mark the points where these lines intersect the horizontal line representing the sea level.

### Step 3: Calculate the Distance to the Ship at 45°
Using the angle of depression of 45°:
- In the right triangle formed, the height (perpendicular) is 100 m, and the angle is 45°.
- Using the tangent function:
\[ \tan(45°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{100}{d_1} \]
Since \(\tan(45°) = 1\):
\[ 1 = \frac{100}{d_1} \implies d_1 = 100 \text{ m} \]

### Step 4: Calculate the Distance to the Ship at 30°
Now, using the angle of depression of 30°:
- In the right triangle formed, the height (perpendicular) is still 100 m, and the angle is 30°.
- Using the tangent function:
\[ \tan(30°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{100}{d_2} \]
Since \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{100}{d_2} \implies d_2 = 100\sqrt{3} \text{ m} \]

### Step 5: Calculate the Distance Travelled by the Ship
The distance travelled by the ship is the difference between the two distances:
\[ \text{Distance travelled} = d_2 - d_1 = 100\sqrt{3} - 100 \]
Factoring out 100:
\[ \text{Distance travelled} = 100(\sqrt{3} - 1) \]

### Step 6: Approximate the Value
Using \(\sqrt{3} \approx 1.732\):
\[ \text{Distance travelled} \approx 100(1.732 - 1) = 100(0.732) \approx 73.2 \text{ m} \]

### Final Answer
The distance travelled by the ship during the period of observation is approximately **73.2 meters**.

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