physics

A golfer takes two putts to sink the ball, one is (81.6 ft, 31.7 degrees N of E) and the other is (3.20 ft, 53.4 degrees W of N). What is the displacement of the single putt that would sink the ball on the first try?

Well, what I did was make a diagram of the two vectors,
I named the first one Alpha and tried to find both of their components.
Alpha-subX (for the x component)
Alpha-subX = 81.6 ft * cos(31.7) = 69.426

Alpha-subY (for the y component)
Alpha-subY = 81.6 ft * sin(31.7) = 42.878

I also did the same thing for the 2nd putt, and named it Beta.

Beta-subX = 3.20 ft cos(53.4) = 1.907
Beta-subY = 3.20 ft sin(53.4) = 2.569

I "think" I'm supposed to combine the x components with each other

Alpha subX + Beta subX
and Alpha subY + Beta subY

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asked by Henry
  1. Yes, exactly BUT THE SECOND ONE was west of north
    x distance = 69.4 -3.2 sin 53.4 = x
    y distance = 42.9 + 3.2 cos 53.4 = y

    d = sqrt (x^2 + y^2)

    tan of angle north of east = y/x

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    posted by Damon
  2. What do you mean, I did indicate on the question that it was west of north?

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    posted by Henry
  3. But going west is NEGATIVE x

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    posted by Damon
  4. Why did you swap Betas trig functions?

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    posted by Henry
  5. This is right:

    Alpha-subX (for the x component)
    Alpha-subX = 81.6 ft * cos(31.7) = 69.426

    Alpha-subY (for the y component)
    Alpha-subY = 81.6 ft * sin(31.7) = 42.878

    This is WRONG

    Beta-subX = 3.20 ft cos(53.4) = 1.907
    Beta-subY = 3.20 ft sin(53.4) = 2.569

    I CLAIM
    Beta-subX = -3.20 ft SIN(53.4) = 1.907
    Beta-subY = 3.20 ft COS(53.4) = 2.569

    Draw a picture

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    posted by Damon
  6. Because the angle 53.4 is left from the Y axis, not up from the X axis.
    They tricked you with the west of north
    that is left of the Y axis
    DRAW the picture.

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    posted by Damon
  7. How do I know which trig function to use?

    I made my second triangle like this
    _____
    \ |
    \ |
    \ |
    \|

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    posted by Henry
  8. Wait that's not how it's supposed to look.

    Ok

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    posted by Henry
  9. copied your result, should be
    I CLAIM
    Beta-subX = -3.20 ft SIN(53.4) = -2.56
    Beta-subY = 3.20 ft COS(53.4) = 1.90

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    posted by Damon
  10. when the angle is in quadrant II
    53.4 degrees left of north
    then east component is - sin
    then north component is + cos

    if you want the angle from the x axis then it is
    90 - 53.4 = 36.6
    then you could use cos and sin like you want because
    sin (90-x) = cos x
    cos (90-x) = sin x :)

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    posted by Damon
  11. Thank you Damon! I think I understand what you meant when I read your post

    My triangle was correct though, yes? I just referred to them wrong?

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    posted by Henry

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