A balloon is 70 feet from an observer which is rising at 15 ft/sec. At 5 seconds after lift off, how fast is the angle of elevation changing?
did you make your sketch ?
let the height be h ft
let the angle be Ø
tanØ = h/70
h = 70tanØ
dh/dt = 70sec^2 Ø dØ/dt ***
when t = 5 sec, h = 75 ft
tanØ = 75/70 = 15/14
sketching a right-angled triangle with that data, shows a hypotenuse of √421
cosØ = 14/√421
cos^2 Ø = 196/421
sec^2 Ø = 421/196
back to ***
15 = 70(421/196) dØ/dt
dØ/dt = 15(196/(70*421) = .09976 rads/sec
check my arithmetic