Question

You walk 55 m to the north, then turn 60° to your right and walk another 45 m. How far are you from where you originally started?
1 : 87 m
2 : 50 m
3 : 94 m
4 : 46 m

Answer 1

Let's assume I walk from the origin $A=(0,0)$ and reach to $B=(0,55)$ then turned around $60^{\circ}$ to reach finally at $C$ which has been shown in the figure bellow,

Now, we have to find the magnitude of $\overrightarrow{w}=\overrightarrow{AC}$ ,

Since we know that distance between any two points $(x_1,y_1)\&(x_2,y_2)$ is given by,

$$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$$

Thus,

$$||\overrightarrow{w}||=\sqrt{(45\sin(60^{\circ})^2+(55+45\cos(60^{\circ})^2}\\ \implies ||\overrightarrow{w}||=86.74\approx 87m$$

Hence the answer will be option 1. i.e $87m$