An aircraft flies 50 km on a bearing of 065 degrees, then 80 km on a bearing of 040 degrees. Calculate the total distance of the aircraft from its starting point.

The aircraft flies $50 \text{ km}$ on a bearing of $065^\circ$.
Let $N_1$ be the North component and $E_1$ be the East component.

$$N_1 = 50 \cos(65^\circ)$$ $$E_1 = 50 \sin(65^\circ)$$

Using a calculator:

$$N_1 \approx 50 \times 0.4226 = 21.13 \text{ km}$$ $$E_1 \approx 50 \times 0.9063 = 45.32 \text{ km}$$

From the end of the first leg, the aircraft flies $80 \text{ km}$ on a bearing of $040^\circ$.
Let $N_2$ be the North component and $E_2$ be the East component.

$$N_2 = 80 \cos(40^\circ)$$ $$E_2 = 80 \sin(40^\circ)$$

Using a calculator:

$$N_2 \approx 80 \times 0.7660 = 61.28 \text{ km}$$ $$E_2 \approx 80 \times 0.6428 = 51.42 \text{ km}$$

Let $N_{total}$ be the total North displacement and $E_{total}$ be the total East displacement.

$$N_{total} = N_1 + N_2 = 21.13 \text{ km} + 61.28 \text{ km} = 82.41 \text{ km}$$ $$E_{total} = E_1 + E_2 = 45.32 \text{ km} + 51.42 \text{ km} = 96.74 \text{ km}$$

The distance from A is the hypotenuse of the right-angled triangle formed by $N_{total}$ and $E_{total}$.

$$\text{Distance} = \sqrt{N_{total}^2 + E_{total}^2}$$ $$\text{Distance} = \sqrt{(82.41)^2 + (96.74)^2}$$ $$\text{Distance} = \sqrt{6791.4081 + 9358.6276}$$ $$\text{Distance} = \sqrt{16150.0357}$$ $$\text{Distance} \approx 127.08 \text{ km}$$

Rounding to one decimal place:

$$\text{Distance} \approx 127.1 \text{ km}$$

Let $\theta$ be the angle measured clockwise from the North line.

$$\tan(\theta) = \frac{E_{total}}{N_{total}}$$ $$\tan(\theta) = \frac{96.74}{82.41}$$ $$\tan(\theta) \approx 1.1739$$ $$\theta = \arctan(1.1739)$$ $$\theta \approx 49.57^\circ$$

Since both $N_{total}$ and $E_{total}$ are positive, the aircraft is in the North-East quadrant, so the bearing is simply this angle.
Rounding to one decimal place and expressing as a 3-digit bearing:

$$\text{Bearing} \approx 049.6^\circ$$

The distance of the aircraft from A is $\boxed{\text{127.1 km}}$ and its bearing from A is $\boxed{\text{049.6}^\circ}$.