I really need help with this!!! Can someone please help me really quick?!

Question

After two minutes of flying, a plane has risen to an altitude of 3500 feet.  The pilot looks back and can still see the end of the runway from where he took off at a 31 degree angle of depression.  How many feet is the airplane from the end of the runway?  Round to the nearest tenth of a foot.

Answer 1

Draw a diagram.  Let point E represent the end of the runway, point P represents the plane, and point A represents the point on the ground directly below the plane.  

 

We then have a right triangle with angle E = 31°,  side EP = x, and side AP = 3500.

 

So, sin31° = 3500/x.

 

           x = 3500/sin31° ≈ 6795.6 ft = 1.3 miles

Answer 2

Hi Caleb, the plane has an altitude of 3500 feet. The angle to the end of the runway is 31 degrees. Form a right triangle with the altitude (vertical height) as one leg and the distance from the vertical leg to the end of the runway as another leg.

 

Then using the tangent of 31 degrees, you can find out how far the plane is from the end of the runway.

 

tan(31) = opposite / adjacent, where opposite represent the distance to the end of the runway and adjacent represents the elevation of the plane.

 

Solve:

 

opposite (distance to the end of the runway) = adjacent (elevation of the plane) * tan(31)

 

opposite = 3500 * tan(41)

 

opposite ≈ 2103 feet.

 

The plane is approximately 2103 from the end of the runway.

 

Questions?

 

 

Answer 3

 

the angle of depression is equal to the angle of elevation

the vertical distance (altitude) is 3500 feet

the distance from the plane to the end of the runway is the hypotenuse of a right triangle where the vertical leg is 3500 feet

sin=opposite/hypotenuse

sin 31º=3500/d where d=distance from the plane to the end of the runway

d=3500/sin 31º

d=3500/0.51503807491

d=6795.614104 feet

d=6795.6 feet