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Right angle trigonometry homework question

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A steel plate has the form of one-fourth of a circle with a radius of 60 centimeters. Two two-centimeter holes are to be drilled in the plate positioned as shown in the figure. Find the coordinates of the center of each hole.


    2. Relevant equations
    I know it's gotta be a simple sin/cos right angle equation, but I've no clue how they came about to getting the 56 centimeter measurement to be the hypotenuse on the solution that I have in my manual.


    3. The attempt at a solution

    I attempted drawing a few right angle triangles that I thought would work, but nothing came from it. If someone could explain how the solution manual was able to get a triangle with 56 as the hypotenuse, I would be able to move forward from there.

    I've attached both the image of the problem, and the image of the first part of the solution worked out.
     

    Attached Files:

    Last edited: Nov 28, 2012
  2. jcsd
  3. Nov 28, 2012 #2

    jedishrfu

    Staff: Mentor

    for the first hole you have a 30-60-90 rt triangle with hyp 56 so you should be able to compute the x1 and y1

    similarly for the x2 y2
     
  4. Nov 28, 2012 #3
    It's because it's the radius of the circle. Every hypotenuse along the circumference of a circle that is measured from the centre is going to be the radius of the circle.
     
  5. Nov 28, 2012 #4

    Mark44

    Staff: Mentor

    It looks to me like the two drilled holes are a radial distance of 56 cm from the center, and at angles of 30° and 60° from the horizontal edge.

    These values aren't calculated - they're part of the given information in the problem. The positions of the two holes are essentially in polar coordinates, and your job is to find the rectangular coordinates of the holes.
     
  6. Nov 28, 2012 #5
    Right, that's basically what the solution is saying, I guess I just can't see how they figured that right triangle/hypotenuse out..

    Edit: Ohhhh...I wasn't thinking of the 56 as a radial measurement. I feel really dumb now. Thanks guys!
     
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