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 answer:

s =  148.6607 km

Step-by-step explanation:

$$\begin{gathered} A=63{\displaystyle \frac{20}{60}}=63+{\displaystyle \frac{20}{60}}={\displaystyle \frac{63\cdot 60+20}{60}}={\displaystyle \frac{3800}{60}}={\displaystyle \frac{190}{3}}\doteq 63.3333\text{ ° } \\ B=153{\displaystyle \frac{20}{60}}=153+{\displaystyle \frac{20}{60}}={\displaystyle \frac{153\cdot 60+20}{60}}={\displaystyle \frac{9200}{60}}={\displaystyle \frac{460}{3}}\doteq 153.3333\text{ ° } \\ a=50\text{ km } \\ b=140\text{ km } \\ {x}_1=a\cdot \cos A\mathrm{°}=a\cdot \cos 63.3333333333\mathrm{°}=50\cdot \cos 63.3333333333\mathrm{°}=50\cdot 0.448799=22.43996\text{ km } \\ {y}_1=a\cdot \sin A\mathrm{°}=a\cdot \sin 63.3333333333\mathrm{°}=50\cdot \sin 63.3333333333\mathrm{°}=50\cdot 0.893633=44.68163\text{ km } \\ {x}_2=b\cdot \cos B\mathrm{°}=b\cdot \cos 153.333333333\mathrm{°}=140\cdot \cos 153.333333333\mathrm{°}=140\cdot (-0.893633)=-125.10857=(-229422\pi \mathrm{/}5761)\text{ km } \\ {y}_2=b\cdot \sin B\mathrm{°}=b\cdot \sin 153.333333333\mathrm{°}=140\cdot \sin 153.333333333\mathrm{°}=140\cdot 0.448799=62.83189\text{ km } \\ x=x_1+x_2=22.44+(-125.1086)\doteq -102.6686\text{ km } \\ y=y_1+y_2=44.6816+62.8319\doteq 107.5135\text{ km } \\ s=\sqrt{x^2+y^2}=\sqrt{(-102.6686{)}^2+107.5135^2}=10\sqrt{221}=148.6607\text{ km} \end{gathered}$$

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