The depth of water in a harbour, \(d\) metres, at \(t\) hours after midnight is modelled by \(d = 6 + 2.5\sin(30t)^\circ\). The diagram shows the graph of t...

In this model the number of hours after midnight is converted into an angle by the multiplier 30, since \(30\times 24=720\), meaning the sine completes two full cycles in a day. That matches the two high tides and two low tides a day shown on the graph. The 6 is the mean depth and the 2.5 is the amplitude, so the depth always lies between 3.5 m and 8.5 m.

(a) At 05 00, \(t=5\), so the angle is \(30\times 5=150^\circ\):

[M1]

Since \(\sin 150^\circ=0.5\), the depth is \(6+2.5\times 0.5=6+1.25=\) 7.25 m cao [A1].

(b) At 20 00, \(t=20\), so the angle is \(30\times 20=600^\circ\). Angles beyond \(360^\circ\) are perfectly acceptable in the calculator, but note \(600^\circ-360^\circ=240^\circ\), which is in the third quadrant where sine is negative:

[M1]

This gives \(6-2.165...=3.8349...\), so the depth is 3.83 m [A1] (3.8349...).

Check both answers against the range 3.5 m to 8.5 m: 7.25 m is above the mean, as expected shortly after a high tide, and 3.83 m is near the low end, close to a low tide. Make sure the calculator is in degree mode, since the formula carries the degree symbol; in radian mode the answers would be meaningless.