A jet leaves a runway whose bearing is N 35°E from the control tower. After flying 4 miles, the jet turns 90° and flies on a bearing of S 55° E for 5 miles the bearing of the jet from the control tower is mathematically given as
N76.675E
What is the bearing of the jet from the control tower?
Generally, the equation for horizontal distance is mathematically given as
sin<AOC=\frac{AC}{OA}
Therefore
sin <AOC=\frac{AC}{5}
AC=4
and
BE=3.6
Total horizontal distance=4+3.6
Total horizontal distance=7.6
For Total vertical distance
cos<AOC=OC/OA
cos<AOC=OC/5
OC=3
and
A.E=4.8
Total vertical distance=3-4.8
Total vertical distance=-1.3=E.C=y
In conclusion, the angle
[tex]\theta=tan{-1} \frac{-1.3}{7.6}[/tex]
[tex]\theta= -13.324[/tex]
Hence
90-13.324=N76.675E
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