Plane II

A plane flew 50 km on a bearing 63degrees20 and the flew on a bearing 153degrees20 for 140km. Find the distance between the starting point and the ending point

Result

x =  148.661 km

Solution:

u=153+20/60−(63+20/60)=90 x=502+1402=10 221≐148.6607≐148.661 kmu=153+20/60-(63+20/60)=90 \ \\ x=\sqrt{ 50^2+140^2 }=10 \ \sqrt{ 221 } \doteq 148.6607 \doteq 148.661 \ \text{km}

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