Posted by Lg on Saturday, June 15, 2013 at 4:38pm.
During high tide the water depth in a harbour is 22 m and during low tide it is 10m(Assume a 12h cycle).
Calculate the times at which the water level is at 18m during the first 24 hours.
My solution:
first I found the cos equation: H(t)=6cos(π/6t)+16
then..
Let π/6t=Θ
18=6cosΘ+16
1816=6cosΘ
Θ=1.230959417
Then I don't know what's next....

Trigonometry..  Lg, Saturday, June 15, 2013 at 6:56pm
H(t)=6\cos(\frac{\pi}{6} t)+16

Trigonometry..  Steve, Saturday, June 15, 2013 at 7:29pm
cosΘ < 0 in QII and QIII
arccos(1/3) = 1.91 = pi1.23
or pi+1.23 = 4.37
So, t = (6/pi) * (1.91 or 4.37)
t = 3.64 or 8.35 hours
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