Four ships A,B,C and D are at sea in the following relative positions: B is on the straight line segment AC, B is due north of D and D is due west of C. The distance B and D is 2km. `angleBDA = 40^(@), angleBCD = 25^(@)`. What is the distance between A and D. (`sin 25^(@) = 0.423`).

Answer

To find the distance between ships A and D, we can use the information given about their positions and the angles involved. Let's break down the solution step by step.

### Step 1: Understand the Positions
- We have four ships: A, B, C, and D.
- B is on the line segment AC.
- B is due north of D.
- D is due west of C.
- The distance between B and D is 2 km.

### Step 2: Draw a Diagram
- Draw a coordinate system where:
- Place D at the origin (0, 0).
- Since B is due north of D, place B at (0, 2) (2 km north).
- Let C be at (x, 0) since D is due west of C.
- A will be somewhere along the line extending from B to C.

### Step 3: Define Angles
- Given:
- Angle BDA = 40° (angle between line BD and line AD).
- Angle BCD = 25° (angle between line BC and line CD).

### Step 4: Use the Law of Sines in Triangle BDC
In triangle BDC, we can use the Law of Sines:
\[ \frac{BD}{\sin(\angle BCD)} = \frac{BC}{\sin(\angle BDC)} \]
Where:
- BD = 2 km
- \(\angle BCD = 25^\circ\)
- \(\angle BDC = 180^\circ - (90^\circ + 25^\circ) = 65^\circ\)

Substituting the known values:
\[ \frac{2}{\sin(25^\circ)} = \frac{BC}{\sin(65^\circ)} \]

### Step 5: Calculate BC
Using the value of \(\sin(25^\circ) = 0.423\):
\[ BC = \frac{2 \cdot \sin(65^\circ)}{\sin(25^\circ)} \]
To find \(\sin(65^\circ)\), we can use the identity \(\sin(65^\circ) = \cos(25^\circ)\):
\[ \cos(25^\circ) = \sqrt{1 - \sin^2(25^\circ)} = \sqrt{1 - (0.423)^2} = \sqrt{1 - 0.1785} = \sqrt{0.8215} \approx 0.906 \]
Now substitute to find BC:
\[ BC = \frac{2 \cdot 0.906}{0.423} \approx \frac{1.812}{0.423} \approx 4.28 \text{ km} \]

### Step 6: Use Triangle ABD
Now, in triangle ABD, we can again use the Law of Sines:
\[ \frac{AD}{\sin(\angle ABD)} = \frac{AB}{\sin(\angle ADB)} \]
Where:
- \(\angle ABD = 40^\circ\)
- \(\angle ADB = 90^\circ - 40^\circ = 50^\circ\)

### Step 7: Find AD
Using the value of AB (which is BC):
\[ AD = \frac{BC \cdot \sin(40^\circ)}{\sin(50^\circ)} \]
Using \(\sin(40^\circ) \approx 0.643\) and \(\sin(50^\circ) \approx 0.766\):
\[ AD = \frac{4.28 \cdot 0.643}{0.766} \approx \frac{2.748}{0.766} \approx 3.59 \text{ km} \]

### Final Answer
The distance between ships A and D is approximately **3.59 km**.