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Math Help - Trigonometric ratio problem

  1. #1
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    Trigonometric ratio problem

    I don't know how to do the questions below↓ Can anyone help me to solve this question? Thanks.
    Trigonometric ratio problem-0001.jpg
    Please help, thx.
    Attached Thumbnails Attached Thumbnails Trigonometric ratio problem-0001.jpg  
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  2. #2
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    Re: Trigonometric ratio problem

    In the first part CD = 250\cos 75

    Therefore the distance AD = \sqrt{ 350^2+ (300+250\cos 75)^2}

    Speed = Distance \div Time = \frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60} metres per second
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  3. #3
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    Re: Trigonometric ratio problem

    Thx a lot.
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  4. #4
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    Re: Trigonometric ratio problem

    Quote Originally Posted by pickslides View Post
    In the first part CD = 250\cos 75 <--- shouldn't that read 250 * cot(75)

    Therefore the distance AD = \sqrt{ 350^2+ (300+250\cos 75)^2}

    Speed = Distance \div Time = \frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60} metres per second
    I really have some difficulties to understand your solution ....
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  5. #5
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    Re: Trigonometric ratio problem

    so what is the correct answer?? Plz help.
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  6. #6
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    Re: Trigonometric ratio problem

    Quote Originally Posted by william24 View Post
    so what is the correct answer?? Plz help.
    Pickslides got |\overline{CD}| \approx 64.705\ m

    I got |\overline{CD}| \approx 66.99\ m

    So the values of the speed differ first at the 3rd decimal. Not much harm done.
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  7. #7
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    Re: Trigonometric ratio problem

    I wanna know the steps and how the steps made, My maths is really poor
    Thx for ur helping, earboth
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  8. #8
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    Re: Trigonometric ratio problem

    The steps are the following. You need to find Dickie's walking speed.

    \mbox{walking speed} = \frac{\mbox{distance travelled}}{\mbox{time travelled}}.

    You know "time travelled", so you only need to find "distance travelled."
    You can apply Pythagoras' theorem to find the distance from D to A if you know AB and BD.
    AB is given (350 meters), and BD = CD + 300 meters, so you need to find CD, which you can derive from the fact that CE as well as the angle at D are given:

    \tan(75) = \frac{EC}{CD},

    so CD\cdot\tan(75) = EC, and CD=\frac{EC}{\tan(75)}=EC\cdot\cot(75)=250\cdot \cot(75).

    Now AD=\sqrt{AD^2+BD^2}=\sqrt{350^2+(300+250\cdot\cot(  75))^2},
    so the speed is \sqrt{350^2+(300+250\cdot\cot(75))^2}/12 meters per minute. (Convert to whatever unit required.)

    Try to understand the steps above, and then solve problem (b) on your own.
    Thanks from william24
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  9. #9
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    Re: Trigonometric ratio problem

    Thx. It's clear
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  10. #10
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    Re: Trigonometric ratio problem

    Quote Originally Posted by tomasklos View Post
    Now AD=\sqrt{AD^2+BD^2}
    Of course, that should be AD=\sqrt{AB^2+BD^2}.
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