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

18-16=-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 = pi-1.23

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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