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 =  148.661 km

Solution:

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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For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
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See also our trigonometric triangle calculator.

 
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