Bearing

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

Correct result:

s =  148.6607 km

Solution:

$$\begin{gathered} A=63+20\mathrm{/}60={\displaystyle \frac{190}{3}}\doteq 63.3333^{\circ } \\ B=153+20\mathrm{/}60={\displaystyle \frac{460}{3}}\doteq 153.3333^{\circ } \\ {x}_1=50\cdot \cos A^{\circ }=50\cdot \cos 63.333333333333^{\circ }=50\cdot 0.448799=22.43996\text{ km } \\ {y}_1=50\cdot \sin A^{\circ }=50\cdot \sin 63.333333333333^{\circ }=50\cdot 0.893633=44.68163\text{ km } \\ {x}_2=140\cdot \cos B^{\circ }=140\cdot \cos 153.33333333333^{\circ }=140\cdot (-0.893633)=-125.10857=(-229422\pi \mathrm{/}5761)\text{ km } \\ {y}_2=140\cdot \sin B^{\circ }=140\cdot \sin 153.33333333333^{\circ }=140\cdot 0.448799=62.83189\text{ km } \\ x=x_1+x_2=22.44+(-125.1086)\doteq -102.6686\text{ km } \\ y=y_1+y_2=44.6816+62.8319\doteq 107.5135\text{ km } \\ s=\sqrt{x^2+y^2}=\sqrt{(-102.6686{)}^2+107.5135^2}=10\sqrt{221}=148.6607\text{ km} \end{gathered}$$

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