An aeroplane has to go from a point P to another point Q, 1000 km away due north. Wind is blowing due east at a speed of `200 km//h.` The air speed of plane is `500 km//h.`
(a) Find the direction in which the pilot should head the plane to reach the point Q.
(b) Find the time taken by the plane to go from P to Q.

Explanation

To solve the problem, we will break it down into two parts: (a) finding the direction in which the pilot should head the plane, and (b) calculating the time taken by the plane to travel from point P to point Q.

### Part (a): Finding the Direction

1. **Understanding the Problem**:
- The plane needs to travel due north to reach point Q, which is 1000 km away.
- The wind is blowing due east at a speed of 200 km/h.
- The airspeed of the plane (the speed of the plane relative to the air) is 500 km/h.

2. **Setting Up the Coordinate System**:
- Let the north direction be the positive y-axis and the east direction be the positive x-axis.

3. **Identifying the Velocity Components**:
- The velocity of the wind (V_wind) is 200 km/h towards the east (x-direction).
- The velocity of the plane (V_plane) is 500 km/h, which we need to resolve into components.

4. **Using Trigonometry**:
- Let θ be the angle the plane makes with the north direction (positive y-axis).
- The northward component of the plane's velocity is V_plane * cos(θ).
- The eastward component of the plane's velocity is V_plane * sin(θ).

5. **Setting Up the Equation**:
- To reach point Q directly north, the eastward component of the plane's velocity must equal the wind's velocity:
\[ V_{plane} \cdot \sin(\theta) = V_{wind} \]
- Substituting the known values:
\[ 500 \cdot \sin(\theta) = 200 \]

6. **Solving for sin(θ)**:
- Rearranging gives:
\[ \sin(\theta) = \frac{200}{500} = 0.4 \]

7. **Finding θ**:
- Taking the inverse sine:
\[ \theta = \sin^{-1}(0.4) \]
- This gives θ ≈ 23.58°.

8. **Direction of the Plane**:
- Since the wind is blowing east, the pilot should head the plane at an angle of approximately 23.58° west of north.

### Part (b): Finding the Time Taken

1. **Calculating the Effective Velocity**:
- The effective velocity of the plane in the northward direction can be calculated using the Pythagorean theorem:
\[ V_{effective} = \sqrt{(V_{plane} \cdot \cos(\theta))^2 + (V_{wind})^2} \]
- First, calculate the northward component:
\[ V_{north} = 500 \cdot \cos(23.58°) \approx 500 \cdot 0.9239 \approx 461.95 \text{ km/h} \]
- The effective velocity is:
\[ V_{effective} = \sqrt{(461.95)^2 + (200)^2} \approx \sqrt{213,000 + 40,000} \approx \sqrt{253,000} \approx 503 \text{ km/h} \]

2. **Calculating Time**:
- The time taken to travel the distance of 1000 km is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Velocity}} = \frac{1000 \text{ km}}{503 \text{ km/h}} \approx 1.987 \text{ hours} \]

### Final Answers:
- (a) The pilot should head approximately 23.58° west of north.
- (b) The time taken by the plane to go from P to Q is approximately 1.99 hours.