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March 26, 2017

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a clock has a dial face of 12 inches radius the minute hand is 9 inches while the hour hand is 6 inches the plane of rotation of the hour hand is 2 inches above the plane of rotation of the minute hand. Find the distance between the tips of the minute and hour hand at 5:40 am.

  • trigonometry - ,

    let the plane of rotation of the minute hand be z=0. So, the location of the tip of the minute hand is

    (9cosθ,9sinθ,0)

    Similarly, the location of the tip of the hour hand is

    (6cos θ/12,6sin θ/12,2)

    at 5:40 am, θ = 2π/3, so θ/12 = π/18

    So, the distance between the tips is

    d^2 = (9cos 2π/3 - 6cos π/18)^2
    + (9sin 2π/3 - 6sin π/18)^2
    + 4

    now just evaluate for d

  • trigonometry - ,

    oops. Forgot all those hours
    θ = 2π*5 + 2π/3 = 10π + 2π/3 = 32π/3
    For the minute hand, that's the same value as 2π/3

    but, θ/12 is now 32π/36 = 8π/9

    which affects its trig function values immensely.

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