# Thread: [SOLVED] harbour depth trig functions

1. ## [SOLVED] harbour depth trig functions

I had a quick search on the forums for help on my question and found this http://www.mathhelpforum.com/math-he...ays-tides.html (also i think i accidently posted this thread in the wrong sub-forum)

However my question is a little different in terms of the placement of "t" and the fact that i'm useless with fractions in equations.

A large ship can only enter the harbour when the depth of water is at
least 1.3 metres above the horizontal mark. At what times during the
week commencing 30 March can the ship enter the harbour?
The equation is $2.5 \sin \left(\frac{\pi t}{12}\right)$

The question in the link is similar but with the placement of the "t" in the equation. I know from my graph and workings that 1.250m occurs at Hours 2, 10, 26 and 34 for 0 ≤ t ≤ 36.

I tried to rearrange the equation for 1.3 by doing:
2.5sin(pi)t/12 = 1.3
sin(pi)t/12 = 0.52

but then i got stuck. Any inputs or hints pls?

2. Originally Posted by PIR
I had a quick search on the forums for help on my question and found this http://www.mathhelpforum.com/math-he...ays-tides.html (also i think i accidently posted this thread in the wrong sub-forum)

However my question is a little different in terms of the placement of "t" and the fact that i'm useless with fractions in equations.

The equation is $2.5 \sin \left(\frac{\pi t}{12}\right)$

The question in the link is similar but with the placement of the "t" in the equation. I know from my graph and workings that 1.250m occurs at Hours 2, 10, 26 and 34 for 0 ≤ t ≤ 36.

I tried to rearrange the equation for 1.3 by doing:
2.5sin(pi)t/12 = 1.3
sin(pi)t/12 = 0.52

but then i got stuck. Any inputs or hints pls?
Use arcsin-function:

$\dfrac{\pi \cdot t}{12}=\arcsin(0.52)\approx 0.546851$

Now solve for t. I've got $t\approx 2.08882$ which correspond to 2 o'clock; 5 min; 20 sec

3. Originally Posted by earboth
Use arcsin-function:

$\dfrac{\pi \cdot t}{12}=\arcsin(0.52)\approx 0.546851$

Now solve for t. I've got $t\approx 2.08882$ which correspond to 2 o'clock; 5 min; 20 sec
I divided, inversed, divided in the end. Worked for me and i got 2.088 too