What is the change in horizontal distance from low tide to high tide?

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Homework Help Overview

The problem involves determining the change in horizontal distance from a cargo ship to a dock as the tide changes, specifically comparing low tide and high tide scenarios. The unloading ramp's angles with the horizontal at these two states are given, along with its length.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of trigonometric ratios to analyze the angles and lengths involved in the problem. There is confusion regarding the application of the ramp length and how it relates to the horizontal distances at different tide levels.

Discussion Status

Some participants have suggested drawing diagrams to visualize the problem, while others have pointed out potential inaccuracies in the representations. There is ongoing exploration of the relationships between the angles and the ramp length, but no consensus has been reached on the correct approach or interpretation.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the trigonometric relationships without providing direct solutions.

Nelo
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Homework Statement


A cargo ship is tied up at the dock. At low tide, a 12-m long unloading ramp slopes down from the ship to the dock and makes an angle of 30 degrees to the horizontal. At high tide, the ship is closer to the dock, and the unloading ramp makes an angle of 45 degrees t othe horizontal.

a) Determine the change in the horizontal distance from the ship to the dock from low tide to high tide. Express the distance as an exact value and as an approximate value


Homework Equations





The Attempt at a Solution



So, looking at the angles i determined that both of these angles are in relation to the exact trig ratio triangles.

I know taht the answer is 6([sqrt3 - [sqrt] 2 ) but I don't know the steps.

I know that the first triangle, which is the one at 60 degrees has a height of [sqrt]3 , and the second one has a slant height of the [sqrt] 2, But where does the 6 come from? and how does the 12-m long unloading ramp come into play in this problem?
 
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anyone?
 
You might want to draw a diagram and use the appropriate trigonometric ratios then.
 
wat...?
 
Did you draw a diagram showing all information in the problem? If so, post it here, and if not, draw one and then post it.
 
Why do you have two diagrams? The situation only needs one. Also I don't think you included the length correctly, see the image attached. AD=BD=Ramp Length
 

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k well, here's a dif one.

If 0º<A<360º

Find all measures of <A

g) Cos A = - [sqr3] / 2

I took the - sqr of 3 /2 and got 150. so that's one possible value.

That point lies somewhere in the second quadrant. If you do 180-150 *since its still in the second quadrant) you get 30. however the answer is 220 for some reason, why is that? why do you do 360 - 150 to get that answer .
 
Cosine is negative in the second and third quadrants. It would be beneficial to note that

cos(90+A)= -sin(A)
cos(180+A)= -cos(A)

The value in third quadrant is not 220. Use the above to figure it out. Also at 30 degrees cosine is positive, not negative.
 
  • #10
210... So you subtracted 360 becasue it is positive in the final quadrant?
 
  • #11
Right.
 
  • #12
Sin A = -0.5

Took the inverse of that and got sin = -30, and since sin is pos in the second quadrant and its moving clockwise i subtracted 270 from that and got 240.. but the answer is 210, 330. how..?
 
  • #13
So when its negetive you add 180 to that negetive value and the get the counterclockwise quadrant. then you start from "all" or quadrant 1 and see where cos becomes positive and subtract from there, yeah?
 
  • #14
If you know a negative value, add 360 since the sine and cosine functions are periodic over 360 degrees. You need to know the form of the sine function to figure out the angles quickly. Since sine goes from 0 to -1 in the third quadrant and from -1 to 0 in the fourth, angles which are equally distant from 180 and 360 have equal sines. For example, 180+30=210 and 360-30=330. Similarly 180+45=225 and 360-45=315 have the same sine.
 
  • #15
Yes, they have the exact same line. So, does it make any difference if let's say i keep a value at sin135 or sin45, aside for showing that i know that these values are equivalent?
 
  • #16
Does that mean cos can have 3 types of values?

Lets say cos is 85. cos85.

then i can do 360-85 and get cos275.

then cos 275-180 =cos95

and if i do -cos95 i can get the exact same value...
 
  • #17
Basically if you draw vertical lines on the unit circle, the cosine is the same, and if you draw horizontal lines, the sine is the same. If you add 360 to an angle, you will get the same value for any trigonometric ratio.
 
  • #18
How do you convert a decimal, like 0.3061 into a radical.

18a) Find the exact value of each of the following.

a) sin 30*sin45*sin60

I got 0.3061 as the decimal, how do i turn that into a radical? I used the calculator to solve, and the exact triangles. :S
 
  • #19
Nelo said:
a) sin 30*sin45*sin60

I got 0.3061 as the decimal...
Interesting. I got about 0.2563. We're evaluating
sin (30 rad) * sin (45 rad) * sin (60 rad)
after all. :rolleyes:

Oh, you mean DEGREES? Then for goodness sake, write in the degree symbol!

Nelo said:
I got 0.3061 as the decimal, how do i turn that into a radical? I used the calculator to solve, and the exact triangles. :S
You don't. You evaluate the 3 sine expressions separately to get their exact values, and then multiply the fractions together.
 
  • #20
You need to know the exact values of sine and cosine at 0,30,45,60 and 90 degrees at least for trigonometry.
 

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