A person on a ship sailing north sees two lighthouses which are 6 km apart, in a line due west . After an hour's tailing one of them bears south west and the other southern south west. The ship is travelling at a rate of

Options

A. 12 km/hr
B. 6 km/hr
C. `3sqrt2` km/hr
D. `(6+3sqrt2)` km/hr

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem and set up the diagram
We have two lighthouses, A and B, which are 6 km apart in a line due west. The ship is initially sailing north and after one hour, the lighthouses are seen at angles of south-west and south-south-west respectively.

### Step 2: Define the positions
- Let the position of the first lighthouse (A) be at point (0, 0).
- The position of the second lighthouse (B) will then be at point (-6, 0) since they are 6 km apart in a line due west.
- Let the position of the ship after one hour be point C.

### Step 3: Determine the angles
- The angle from the ship to lighthouse A (south-west) is 45 degrees.
- The angle from the ship to lighthouse B (south-south-west) is 22.5 degrees.

### Step 4: Set up the triangle
Using the angles and the distance between the lighthouses, we can set up the triangle ADB where:
- A is the position of lighthouse A
- B is the position of lighthouse B
- D is the position of the ship after one hour (point C)

### Step 5: Use trigonometry to find distances
Using the tangent function:
- For triangle ACD (where D is the position of the ship after one hour):
\[ \tan(45^\circ) = \frac{AD}{CD} \]
Since \(\tan(45^\circ) = 1\), we have:
\[ AD = CD \]

- For triangle BCD:
\[ \tan(22.5^\circ) = \frac{BD}{CD} \]
Rearranging gives us:
\[ BD = CD \cdot \tan(22.5^\circ) \]

### Step 6: Calculate distances
Since the distance between A and B is 6 km:
\[ AB = AD + BD = 6 \]
Substituting the expressions we derived:
\[ CD + CD \cdot \tan(22.5^\circ) = 6 \]
\[ CD(1 + \tan(22.5^\circ)) = 6 \]
Thus:
\[ CD = \frac{6}{1 + \tan(22.5^\circ)} \]

### Step 7: Calculate the speed of the ship
Since the ship traveled for one hour, the speed of the ship is equal to the distance traveled:
\[ \text{Speed} = CD \]
Substituting the value of \(CD\):
\[ \text{Speed} = \frac{6}{1 + \tan(22.5^\circ)} \]

### Step 8: Final calculation
Using the known value of \(\tan(22.5^\circ) = \sqrt{2} - 1\):
\[ \text{Speed} = \frac{6}{1 + (\sqrt{2} - 1)} = \frac{6}{\sqrt{2}} = 6\sqrt{2}/2 = 3\sqrt{2} \text{ km/h} \]

### Conclusion
The speed of the ship is \(3\sqrt{2} \text{ km/h}\).

---