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 need to break it down into two parts as specified in the question.

### Part (a): Find the direction in which the pilot should head the plane to reach point Q.

1. **Identify the given information:**
- Distance from P to Q (D) = 1000 km (due North)
- Wind speed (Vw) = 200 km/h (due East)
- Air speed of the plane (Va) = 500 km/h

2. **Set up the components of the velocities:**
- The plane's velocity can be broken down into two components:
- Northward component: \( Va \cos(\theta) \)
- Eastward component: \( Va \sin(\theta) \)

3. **Condition for reaching point Q:**
- To ensure the plane reaches point Q directly north, the eastward component of the plane's velocity must cancel out the wind's velocity:
\[ Va \sin(\theta) = Vw \]
Substituting the values:
\[ 500 \sin(\theta) = 200 \]
\[ \sin(\theta) = \frac{200}{500} = \frac{2}{5} \]

4. **Calculate the angle \( \theta \):**
- To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}\left(\frac{2}{5}\right) \]

5. **Find the cosine of the angle:**
- Using the Pythagorean identity:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{2}{5}\right)^2} = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \]

6. **Determine the heading direction:**
- The pilot should head the plane at an angle \( \theta \) north of west, where \( \theta = \sin^{-1}\left(\frac{2}{5}\right) \).

### Part (b): Find the time taken by the plane to go from P to Q.

1. **Calculate the northward component of the plane's velocity:**
\[ Va \cos(\theta) = 500 \cos(\theta) = 500 \cdot \frac{\sqrt{21}}{5} = 100\sqrt{21} \text{ km/h} \]

2. **Use the formula for time:**
\[ \text{Time} (T) = \frac{\text{Distance}}{\text{Velocity}} = \frac{D}{Va \cos(\theta)} \]
Substituting the values:
\[ T = \frac{1000}{100\sqrt{21}} = \frac{10}{\sqrt{21}} \text{ hours} \]

### Final Answers:
- **(a)** The pilot should head the plane at an angle \( \theta = \sin^{-1}\left(\frac{2}{5}\right) \) north of west.
- **(b)** The time taken by the plane to go from P to Q is \( \frac{10}{\sqrt{21}} \) hours.