Answer:
The horizontal distance from the plane to the person on the runway is 20408.16 ft.
Step-by-step explanation:
Consider the figure below,
Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°
BC is the horizontal distance from the plane to the person on the runway.
We have to find distance BC,
Using trigonometric ratio,
Here, 
,Perpendicular AB = 4000
Solving for BC, we get,
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft
Use Law of Cosines g^2 = f^2 + h^2 -2fhCosG f^2 = g^2 + h^2 -2ghCosF h^2 = f^2 + g^2 -2fgCosH
f^2 = 28^2 + 15^2 -2*28*15Cos87 28^2 = 31^2 + 15^2 -2*31*15CosG
f^2 = 784 + 225 - 43.96 784 = 961+225 - 930CosG
f^2 = 965.0378 784 - 1186 = -930CosG
f = 31 -402 = -930CosG Divide by -930
.432258 = CosG
Cos^-1(.432258) = G
G = 64 degrees
Angle H = 180 - 64 - 87 = 29 degrees
Side f = 31 Angle F = 87 degrees
Side g = 28 Angle G = 64 degrees
Side h = 15 Angle H =29 degrees
Comment is correct, I mis read. it is figure B since not gorilla is inside not ape
Wesley used 3/4 pounds of nails
1/2=2/4
4/4-1/4=3/4
