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  1. #1
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    solve

    I need some help with these.

    Solve the following

    cos-1(x) = 60 23' 45"
    sin-1 (x) = 60 23' 45"
    tan-1 (x) = 60 23' 45"
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  2. #2
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    Your calculator should have a cos/sin/tan^-1 function. If not you can use google.

    Remember, there are sixty seconds in a minute and sixty minutes in a degree, so when entering it into your calculator, you want to do ((23)(60) + 45)/(3600) Find that decimal, put it after the sixty and find the inverse cos/sin/tan from there.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Wildman View Post
    I need some help with these.

    Solve the following

    [1] cos-1(x) = 60 23' 45"
    [2] sin-1 (x) = 60 23' 45"
    [3] tan-1 (x) = 60 23' 45"
    Hi Wildman,

    I hope you intended these to be 3 separate exercises.

    You are taking the inverse cos, inverse sin, and inverse tan of some ratio to arrive at 60^{\circ}23'45"

    Convert: 60^{\circ}23'45"\Longrightarrow 60+\frac{23}{60}+\frac{45}{3600}=\left(\frac{2899}  {48}\right)^{\circ}

    Now just take the Cos, Sin, and Tan of that amount to find the value of x in each case.

    [1] \cos\left(\frac{2899}{48}\right)^{\circ}\approx .4940

    [2] \sin\left(\frac{2899}{48}\right)^{\circ}\approx .8695

    [3] \tan\left(\frac{2899}{48}\right)^{\circ}\approx 1.7600
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