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The angle of elevation to an airplane viewed from the control tower at an airport is 15°. The tower is 100 feet high and Pilot Cory reports that the altitude of the airplane is 5100 feet (5000 feet higher than the control tower). How far away (ground distance) from the control tower is the airplane?

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kismatkaucha0

Answer:

  • Given: Angle of elevation 15°
  • AB (height of tower)=100
  • AE=?
  • CE = 5100 (5000 feet heigher than the tower)

Now,

tan15°=

[tex] \frac{5100}{ae} [/tex]

ae=

[tex] \frac{5100}{tan15} [/tex]

.: ae = 19033.45

.: the aeroplane is 19033.45 m away from the control tower.

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