An aircraft flies at `400 km//h` in still air. A wind of `200sqrt2 km//h` is blowing from the south towards north. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if `AB=1000 km.`

Explanation

To solve the problem step by step, we will analyze the situation involving the aircraft, the wind, and the desired direction of travel.

### Step 1: Understand the Problem
The aircraft flies at a speed of 400 km/h in still air. There is a wind blowing from the south to the north at a speed of \(200\sqrt{2}\) km/h. The pilot wants to travel from point A to point B, which is located northeast of A, a distance of 1000 km.

### Step 2: Set Up the Vectors
1. **Aircraft's Velocity (V_A)**: The speed of the aircraft is \(400\) km/h.
2. **Wind's Velocity (V_W)**: The wind's speed is \(200\sqrt{2}\) km/h, directed north.
3. **Resultant Velocity (V_R)**: The resultant velocity of the aircraft (V_R) must point northeast.

### Step 3: Determine the Angles
Since northeast is at a \(45^\circ\) angle from the north, we can denote:
- \(V_R\) (resultant velocity) makes an angle of \(45^\circ\) with the north.

### Step 4: Apply the Law of Sines
Using the law of sines in triangle ABC, where:
- \(C\) is the aircraft's direction,
- \(A\) is the wind's direction,
- \(B\) is the resultant direction towards point B.

We have:
\[ \frac{CB}{\sin(45^\circ)} = \frac{AC}{\sin(\alpha)} \]
Where:
- \(CB = 400\) km/h (aircraft speed),
- \(AC = 200\sqrt{2}\) km/h (wind speed),
- \(\alpha\) is the angle the aircraft must steer.

### Step 5: Solve for \(\alpha\)
From the sine law:
\[ \frac{400}{\sin(45^\circ)} = \frac{200\sqrt{2}}{\sin(\alpha)} \]
Substituting \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ \frac{400}{\frac{1}{\sqrt{2}}} = \frac{200\sqrt{2}}{\sin(\alpha)} \]
This simplifies to:
\[ 400\sqrt{2} = \frac{200\sqrt{2}}{\sin(\alpha)} \]
Cross-multiplying gives:
\[ 400\sqrt{2} \sin(\alpha) = 200\sqrt{2} \]
Dividing both sides by \(200\sqrt{2}\):
\[ 2\sin(\alpha) = 1 \implies \sin(\alpha) = \frac{1}{2} \]
Thus, \(\alpha = 30^\circ\).

### Step 6: Determine the Steering Direction
The pilot must steer at an angle of:
\[ 45^\circ + \alpha = 45^\circ + 30^\circ = 75^\circ \]
This means the aircraft should be steered \(75^\circ\) from north towards east.

### Step 7: Calculate the Time of Journey
To find the time taken to travel from A to B:
1. Calculate the magnitude of the resultant velocity \(V_A\) using the law of sines again:
\[ \frac{V_A}{\sin(105^\circ)} = \frac{400}{\sin(45^\circ)} \]
Where \(105^\circ = 180^\circ - 45^\circ - 30^\circ\).

Calculating:
\[ V_A = \frac{400 \cdot \sin(105^\circ)}{\sin(45^\circ)} \]
Using \(\sin(105^\circ) \approx 0.9659\) and \(\sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071\):
\[ V_A = \frac{400 \cdot 0.9659}{0.7071} \approx 546.47 \text{ km/h} \]

2. Finally, calculate the time:
\[ \text{Time} = \frac{\text{Distance}}{\text{Velocity}} = \frac{1000 \text{ km}}{546.47 \text{ km/h}} \approx 1.83 \text{ hours} \]

### Final Answer
The pilot must steer at an angle of \(75^\circ\) from north towards east, and the time of the journey will be approximately \(1.83\) hours.