An airplane takes off 3000ft in front of a 75ft y'all building. At which angle of elevation must the plane take off in order to avoid hitting the building?

Airplane plane should take off at angle of elevation of 1.43°.

Step-by-step explanation:

As given in the figure attached,

Distance of the building from the point of take off (BC) = 3000 ft

Height of the building (AB) = 75 feet

We have to calculate angle of elevation (θ) so that airplane doesn't hit the building.

tanθ =

tanθ =

tanθ =

θ =

θ = 1.432° ≈ 1.43°

Therefore, airplane should take off at angle of elevation of 1.43° to avoid the accident.

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation: