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 =  148.661 km


A=63+20/60=190363.3333  B=153+20/60=4603153.3333  x1=50 cosA=50 cos63.3333333333 =50 0.448799=22.43996 y1=50 sinA=50 sin63.3333333333 =50 0.893633=44.68163 x2=140 cosB=140 cos153.333333333 =140 (0.893633)=125.10857=(229422π/5761) y2=140 sinB=140 sin153.333333333 =140 0.448799=62.83189 x=x1+x2=22.44+(125.1086)102.6686 km y=y1+y2=44.6816+62.8319107.5135 km s=x2+y2=(102.6686)2+107.51352=10 221=148.661 kmA=63+20/60=\dfrac{ 190 }{ 3 } \doteq 63.3333 \ ^\circ \ \\ B=153+20/60=\dfrac{ 460 }{ 3 } \doteq 153.3333 \ ^\circ \ \\ x_{1}=50 \cdot \ \cos A ^\circ =50 \cdot \ \cos 63.3333333333^\circ \ =50 \cdot \ 0.448799=22.43996 \ \\ y_{1}=50 \cdot \ \sin A ^\circ =50 \cdot \ \sin 63.3333333333^\circ \ =50 \cdot \ 0.893633=44.68163 \ \\ x_{2}=140 \cdot \ \cos B ^\circ =140 \cdot \ \cos 153.333333333^\circ \ =140 \cdot \ (-0.893633)=-125.10857=(-229422π/5761) \ \\ y_{2}=140 \cdot \ \sin B ^\circ =140 \cdot \ \sin 153.333333333^\circ \ =140 \cdot \ 0.448799=62.83189 \ \\ 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 }=148.661 \ \text{km}

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