A survey drone takes off and flies 120 m north, then 50 m west, then 70 m at 40° south of east…

Step 1. The three flights. North: $(0, 120)$. West: $(-50, 0)$. At $40^\circ$ south of east: $(70\cos 40^\circ, -70\sin 40^\circ) = (53.6, -45.0)$.

Step 2. Add. $R_x = 0\ \text{m} - 50\ \text{m} + 53.6\ \text{m} = 3.6\ \text{m}$ and $R_y = 120\ \text{m} + 0\ \text{m} - 45.0\ \text{m} = 75.0\ \text{m}$.

Step 3. Combine. $R = \sqrt{(3.6\ \text{m})^2 + (75.0\ \text{m})^2} = 75.1\ \text{m}$. The direction is $\tan^{-1}\dfrac{3.6\ \text{m}}{75.0\ \text{m}} = 2.8^\circ$ east of north.