180mph = 3 mi/min
using x,y,z coordinates,
Stacey is at (-5,0,0)
At time t minutes,
The plane is at (0,3t,3tan15° t) = (0,3t,0.8t)
So, at time t, the angle θ from Stacey to the point under the plane, is
tanθ = 3t/5
So at 12 minutes, tanθ = 36/5
the angle of elevation θ from Stacey to the plane is
tanθ = 9.6/5
the distance at time t is
d^2 = 5^2+(3t)^2+(.8t)^2
so at t=12,
d^2 = 25+1296+92.16
d = 37.6 miles
stacey is standing 5 miles west of the control tower at an airport. A plane took off traveling directly north of the control tower at a rate of 180 mph along the ground at a constant angle of 15 degrees.
A)after 12 minutes, what is the angel between Stacey and the plane along the ground?
B)after 12 minutes, what is the angle of elevation between Stacey and the plane?
C) how far is the plane from Stacey?
1 answer
1 answer