1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Word problem, trig functions

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    3. 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 dont 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?
  2. jcsd
  3. Jul 21, 2011 #2
  4. Jul 21, 2011 #3
    You might want to draw a diagram and use the appropriate trigonometric ratios then.
  5. Jul 21, 2011 #4
  6. Jul 21, 2011 #5
    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.
  7. Jul 21, 2011 #6
  8. Jul 21, 2011 #7
    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

    Attached Files:

  9. Jul 21, 2011 #8
    k well, heres 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 thats 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 .
  10. Jul 21, 2011 #9
    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.
  11. Jul 21, 2011 #10
    210... So you subtracted 360 becasue it is positive in the final quadrant?
  12. Jul 21, 2011 #11
  13. Jul 21, 2011 #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..?
  14. Jul 21, 2011 #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?
  15. Jul 21, 2011 #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.
  16. Jul 21, 2011 #15
    Yes, they have the exact same line. So, does it make any difference if lets say i keep a value at sin135 or sin45, aside for showing that i know that these values are equivelent?
  17. Jul 21, 2011 #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...
  18. Jul 21, 2011 #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.
  19. Jul 21, 2011 #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
  20. Jul 22, 2011 #19


    User Avatar
    Homework Helper

    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!

    You don't. You evaluate the 3 sine expressions separately to get their exact values, and then multiply the fractions together.
  21. Jul 22, 2011 #20
    You need to know the exact values of sine and cosine at 0,30,45,60 and 90 degrees at least for trigonometry.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Word problem, trig functions
  1. Trig word problem (Replies: 1)

  2. Trig word problem (Replies: 2)

  3. Trig word problems (Replies: 2)