Plane II

A plane flew 50 km on a bearing of 63°20' and then flew in the direction of 153°20' for 140km. Find the distance between the starting point and the ending point.

Correct answer:

x =  148.6607 km

Step-by-step explanation:

$$\begin{gathered} \alpha =63°20^{\prime }=63°+\frac{20}{60}°=63.3333°\doteq 63.3333 \\ \beta =153°20^{\prime }=153°+\frac{20}{60}°=153.3333°\doteq 153.3333 \\ \gamma =\beta -\alpha =153.3333-63.3333=90{}^{\circ } \\ a=50\text{km} \\ b=140\text{km} \\ x=\sqrt{a^2+b^2}=\sqrt{50^2+140^2}=10\sqrt{221}=148.6607\text{km} \end{gathered}$$

Try another example