Question

An air force pilot must descend 1500 feet over a distance of 9000 feet to land smoothly on an aircraft carrier. What is the plane's angle of descent?

Answer 1

using the tan formula, tan angle = 1500/9000 solve from here please

Answer 2

Let's name this angle X.\[\angle X=\tan^{-1}{\frac{1500}{900}}\]

Answer 3

@clamin

Answer 4

Oh, there was another solution up... sorry, didn't notice that.

Answer 5

i got .167

Answer 6

in degrees?

Answer 7

you should get about 59 degrees

Answer 8

you saaid 1500/9000

Answer 9

Wait, did you use my formula or Crash's solution -

Answer 10

Yeah, you have to continue on with the arctan.

Answer 11

the methods are the same

Answer 12

idk how u got 59

Answer 13

maybe u were in radians
change it to degrees

Answer 14

i did

Answer 15

Don't you have to arctan the tan fraction to get the degrees?

Answer 16

you have to use inverse tan, not tan

Answer 17

That's what I said ._.;
Unless arctan doesn't mean inv tan. *googles*

Answer 18

Yup, arctan is inv tan.

Answer 19

i know how you got 59..you divide 1500/900 instead of 1500/9000

Answer 20

I got 9.46, and I double-checked to see that I'm in degree mode ._.;

Answer 21

woops my bad

Answer 22

I just found an answer key and it says the answer's 9.5 degrees...

Answer 23

http://prntscr.com/6rvdpd

Answer 24

1500/9000=0.1667
take atan(0.1667)=9.4 deg

Answer 25

9.5.\[\tan^{-1}\frac{1500}{9000}=9.462322208\approx 9.5\]

Answer 26

.165 rad

Answer 27

^ yes.

Answer 28

Don't forget to close the question if you don't need anymore help :)