Question #253755

Two airplanes leave an airport at the same
time. The velocity of the first airplane is
710 m/h at a heading of 20.7
à ƒ ƒ ¢ — ¦
. The velocity
of the second is 560 m/h at a heading of 107à ƒ ƒ ¢ — ¦
.
How far apart are they after 1.8 h?

Expert's answer

Find the angle between the bearings of the airplanes:


θ=107°−20.7°=86.3°.\theta=107°-20.7°=86.3°.


Apply the law of cosines:


D=(v1t)2+(v2t)2−2v1v2t2cos⁡θ=1576 miles.D=\sqrt{(v_1t)^2+(v_2t)^2-2v_1v_2t^2\cos\theta}=1576\text{ miles}.




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