Then you take cos^-1 (0.833)
= 33.65/180 x 23 = 4.3
From where did you get 23?
Actually 33.65 = t*π/6 = t*180/6.
Now find t.
Hey guys just asking a question about transformations of trigonometric graphs
h = Hcos (tπ/6)
Find t when h = 2.5
H = 3 (height of the high tide in meters)
T = 12
a high tide of 3m occurs at 10am
so I went
2.5 = 3cos (tπ/6)
2.5/3 = cos (tπ/6)
0.833= cos (tπ/6)
Then you take cos^-1 (0.833)
= 33.65/180 x 23 = 4.3 (to convert to hours)
4.3 x 6 = tπ
25.8/π = t
t = 8.21
Yeah I think I'm doing it wrong =(
Thanks for any help.
If you intended to convert from degrees to radians, then it is not "23"! Of course, once you have you can just cancel the " s": so that .
But since you knew the argument of cosine was in radians (the only time you use degrees is in problems specifically dealing with triangles in which the angles are given in degrees) it would be simpler to set your calculator to "degree" mode.
4.3 x 6 = tπ
25.8/π = t
t = 8.21
Yeah I think I'm doing it wrong =(
Thanks for any help.
Hey guys thanks for your help
Ok, I'm going to write out the whole question to prevent any confusions (i apologize for any inconveniences):
3. The height of the sea water (due to tides) above the mean sea level is given by the formula h = Hcos (tπ/6), where t is the time in hours after high tide and H is the height of the high tide in metres. Suppose a high tide of 3.0 metres occurs at 10am.
c. Find the times of the day when the height of the tide is 2.5m
so by doing your calculation t = 33.64/30 = 1.12 ... its in radians but don't I need to convert it to hours to get the times of day? That's why I multiplied by 33.65/180 x 23 = 4.3 hours