Ur solving it like physics solve it like mathematics

The aircraft flies at 20 km/hr for 3 hours, covering a distance $d_1 = 20 \text{ km/hr} \times 3 \text{ hr} = 60 \text{ km}$.
The bearing is $052^\circ$. The coordinates $(x_1, y_1)$ are calculated using sine for the East component and cosine for the North component:

$$x_1 = d_1 \sin(52^\circ) = 60 \sin(52^\circ)$$ $$y_1 = d_1 \cos(52^\circ) = 60 \cos(52^\circ)$$ $$x_1 \approx 60 \times 0.78801 = 47.2806 \text{ km}$$ $$y_1 \approx 60 \times 0.61566 = 36.9396 \text{ km}$$

So, the position after the first leg is $P_1 = (47.2806, 36.9396)$.

The aircraft flies at 30 km/hr for 1.5 hours, covering a distance $d_2 = 30 \text{ km/hr} \times 1.5 \text{ hr} = 45 \text{ km}$.
The bearing is $028^\circ$. The displacement components $(\Delta x_2, \Delta y_2)$ for this leg are:

$$\Delta x_2 = d_2 \sin(28^\circ) = 45 \sin(28^\circ)$$ $$\Delta y_2 = d_2 \cos(28^\circ) = 45 \cos(28^\circ)$$ $$\Delta x_2 \approx 45 \times 0.46947 = 21.1262 \text{ km}$$ $$\Delta y_2 \approx 45 \times 0.88295 = 39.7328 \text{ km}$$

The final position $(x_f, y_f)$ is the sum of the coordinates from the first leg and the displacements from the second leg:

$$x_f = x_1 + \Delta x_2 = 47.2806 + 21.1262 = 68.4068 \text{ km}$$ $$y_f = y_1 + \Delta y_2 = 36.9396 + 39.7328 = 76.6724 \text{ km}$$

So, the final position of the aircraft is $(68.4068, 76.6724)$.

The distance from the origin $(0,0)$ to the final position $(x_f, y_f)$ is found using the distance formula:

$$D = \sqrt{x_f^2 + y_f^2}$$ $$D = \sqrt{(68.4068)^2 + (76.6724)^2}$$ $$D = \sqrt{4679.49 + 5878.62}$$ $$D = \sqrt{10558.11}$$ $$D \approx \boxed{\text{102.75 km}}$$

2) The bearing of the aircraft from the airstrip

The bearing $\theta$ is the angle measured clockwise from the North axis (positive y-axis). Since both $x_f$ (East) and $y_f$ (North) are positive, the aircraft is in the first quadrant (North-East).

$$\tan(\theta) = \frac{\text{East displacement}}{\text{North displacement}} = \frac{x_f}{y_f}$$ $$\tan(\theta) = \frac{68.4068}{76.6724}$$ $$\tan(\theta) \approx 0.89210$$ $$\theta = \arctan(0.89210)$$ $$\theta \approx 41.72^\circ$$

Expressed as a three-digit bearing:

$$\text{Bearing} = \boxed{\text{041.7°}}$$