# trigonometry question

Discussion in 'General Math' started by Jason, Nov 17, 2005.

1. ### JasonGuest

Hello, I am having a problem with a trig question maybe someone can
give me a clue where I am going wrong!

The height h in metres of water in a harbour above low tide is given
by h = 14 - 10cos(30t), where t is hours since midday. A ship can enter
the harbour when the height of the water is > 20 metres. Between what
times is it safe for boats to enter the harbour?

Okay, so this is what I tried:

I think I need to find two solutions where the value of t makes h 20,
and then format the answer as two times?

so I can rewrite it as:

20 = 14 - 10cos(30t) - solve when h = 20
20 - 14 = 10cos(30t) - move the 14 over
(20 - 14) / 10 = cos(30t) - move the coefficient over
30t = cos^-1((20 - 14) / 10) - use arccosine to get t out
t = cos^-1((20 - 14) / 10) / 30 - move the coefficient over

so two possible solutions to t would be +/- 1.77?

so midday - 1.77 hours, and midday + 1.77 hours

boat can pass between 10:14am and 1:46pm?
but apparently this answer is wrong and I have to do it again, but I'm
very stuck!

please point me in the right direction!

thankyou for any suggestions.

Jason, Nov 17, 2005

2. ### William ElliotGuest

If arccos is correct, then at those two times
the h is 20 and the boat can't pass because
it's necessary h > 20, not h = 20.
What can you say about h before and after arccos 6/10 ?
You want to know when f(t) = h - 20 is positive,
ie all values of t when f(t) is positive.

William Elliot, Nov 17, 2005

3. ### a1jrjGuest

Stop right there - you have dropped a sign.

Looking at the equation, h can only be bigger than 14 if cos(30t) is
negative.
Where is cos negative? hint: cos(-x) is cos(x) not as you seem to think
-cos(x).

Should be plain sailing from there on ;-)
Incidentally, drawing a graph to check trig answers is beneficial at
least initially.

HTH
JJ

a1jrj, Nov 17, 2005
4. ### [Mr.] Lynn KurtzGuest

Just curious, why do you carry (20 - 14) / 10) all the way through
your work without simplifying it? If nothing else, you could save

--Lynn

[Mr.] Lynn Kurtz, Nov 17, 2005
5. ### ProginoskesGuest

He probably got yelled at for using his calculator to find e^2, and
he's overcompensating for that mistake.

--- Christopher Heckman

Proginoskes, Nov 18, 2005
6. ### JasonGuest

I find it easier to backtrack later if i make a mistake, i would
simplify if it got more complicated than this.

Thanks to everyone who responded, I think I've got the right answer
now!

Thanks, Jason.

Jason, Nov 18, 2005
7. ### Grover HughesGuest

^ sign missing here!

20 - 14 = -10cos(30t), not +10 etc.

you missed the minus sign! Try it again after this correction and you should
be OK. Best wishes,

gwh

Grover Hughes, Nov 19, 2005
8. ### ticbolGuest

Jason, Nov 17, :
by h = 14 - 10cos(30t), where t is hours since midday. A ship can enter

the harbour when the height of the water is > 20 metres. Between what
times is it safe for boats to enter the harbour?
and then format the answer as two times?
You transposed the 14 to the lefthand side, allright, but why was the
"10cos(30t)" changed from negative to positive at the righthand side?
It should have remained as negative. So,

6 = -10cos(30t)
cos(30t) = -6/10 = -0.60 -----negative cosine, meaning the angle 30t
is in the 2nd or 3rd quadrants.

30t = arccos(-0.6)
30t = 126.87 ------in the 2nd quadrant
30t = (180 -126.87) +180 = 233.13 ------in the 3rd quadrant
Hence,
t = 126.87 /30 = 4.23 -------4 +0.23(60) = 4hr +13.8min = 4:13.8 PM
t = 233.13 /30 = 7.77 -------7 +0.77(60) = 7 +46.2 = 7:46.2 PM

That means, the ships can safely enter the harbor from 4:14 PM to 7:46

---------
If you want, if you know how to do it, you can draw the graph of
h = 14 -10cos(30t)
on t,h cartesian axes---t being horizontal axis, in hrs; h being
vertical axis, in meters; (0,0) being (12noon,0m).
It is a sinusoidal curve, a cosine curve, with a period of 360/30 = 12
hrs, an amplitude of 10m, and it was shifted vertically up at 14m.
Max h = 14 +10 = 24m
Min h = 14 -10 = 4m.
h at 12:00Noon = 14 -10cos(30*0) = 14 -10 = 4m.

Then draw a horizontal line at h=20m.
You'd see that this h=20m line cuts the graph at two points---one point
after the (1/4)period or 3:00PM, and the other point before the
(3/4)period or 9:00PM.

------
Your "The height h ?in metres of water in a harbour above low tide is
given
by h = 14 - 10cos(30t),..." is misleading.

The "...above low tide..." should have not been there----because it was
found out that at low tide, h=4m.
At low tide does not mean there is no water at the harbor.

ticbol, Nov 20, 2005