I really need help with this math problem... can you do it?

Question

An air force pilot must descend 1500ft. over a distance of 9000 ft.

What is the plane's angle of descent?

Use Trig. Ratios (Sin, Cosine, Tangent)

Please show all steps and explain... thanks!

Answer 1

the graph is a right angled triangle:

high= 1500

lenght= 9000

sin x = opposite side / hypotenus

you must calculate the hypotenus

a^2 = b^2 + c^2

being "a" the hypotenus; both b (high) and c(lenght) the sides

Then:

sin x = 1500 / a

x = arc sin (1500/a)

x is the angle asked

Answer 2

sorry

Answer 3

sin(x) = 1500/9000 = 0.16666666666666666666666666666667, where x is the angle.

think  of it as a triangle.

the 9000 ft form the hypotenuse, and the 1500 a cathetus (the smaller one).

the angle of descent is the one formed by the second cathetus ( the one different of 1500 ft), and the hypotenuse.

as you know, sin (x) = the opposite cahetus / hypotenuse.

use a scientific calculator, to calculate arcsin(0.1666666666666666666666666666666... x ( i.e. the angle)

Answer 4

the direction the plane travels describes a triangle.  ( like I\ )

the left, vertical side is 1500 ft.

the hypotenuse is 9000 ft.

you want the angle between them

in a right triangle the relationship between the opposite and the hypotenuse is sin.  to get the angle from the ground up to 1500 ft along the 9000 ft path of the plane you take the inverse sign of 1500/9000

sin^-1(1500/9000) = 9.6 degrees [remember to put your calculator in degrees]

by like triangles that is 9.6 degrees below horizontal or 99.6 degrees south of perpendicular to the earths surface.