Introduction to Circular Functions help?!

The depth of water at the entrance of the harbour t hours after high tide is D metres, where D=p+q cos(rt) for suitable constants p,q,r. At high tide the depth is 7m; at low tide, 6 hours later, the depth is 3m.

Show that r =30?

Can someone explain how I can possibly figure this out?
Thankyou

in D = p + q cos (rt)

the period is 2pi / r, or r = 2pi / period

since the difference between low tide and hi tide is 6 hours, that means the period form high tide to high tide (or low tide to low tide) is 12 hours

r = 2pi / 12 = pi / 6

or, if you're working in degrees, period = 360 / r , meaning that r = 360 / period
r = 360 / 12 = 30
here, the angle will be given in degrees if r = 30

you didn't ask, but p = (max + min) / 2 ==> (7 + 3) / 2 = 5
and q = (max - min) / 2 ==> (7 - 3) / 2 = 2

for this problem, D(t) = 5 + 2cos (30t), where the angle is in degrees
check:
at t = 0, D(0) = 5 + 2cos (0) = 5 + 2 = 7
at t = 6, D(6) = 5 + 2cos (180) = 5 - 2 = 3
at t = 12, D(12) = 5 + 2cos (360) = 5 + 2 = 7, back to high tide 12 hours later...