Question

A man goes 80 m due east and then 150 m due north. How far is he from the starting point?

Answer 1

Let the man starts from point A and goes 80 m due east to B.
Then, from B, he goes 150 m due north to C.



We need to find AC.
In right-angled triangle ABC, we have:

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ AC=\sqrt{{80}^2+{150}^2} \\ =\sqrt{6400+22500} \\ =\sqrt{28900} \\ =170\mathrm{m} \end{gathered}$$

Hence, the man is 170 m away from the starting point.