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Math Help - Help Please!

  1. #1
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    Help Please!

    Really stuck with these.. please help!

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  2. #2
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    Hello tristan1,
    I presume that your triangles are rectangular.

    So use the equations which define the sin- and cos-function:

    \sin{52^\circ}=\frac{32}{x}

    \cos{23^\circ}=\frac{5.2}{x}

    \sin{71^\circ}=\frac{162}{x}

    \sin{71^\circ}=\frac{431}{x}

    I guessed the last number only, so may be you've to repair my equation a little bit.

    To solve the equations should be of no trouble for you.

    Bye
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  3. #3
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    Thanks and they are rectangular
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  4. #4
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    The correct english for this type of triangle is "right triangle" which implies 1 of the angles is 90 degrees.

    Thanks Earboth
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  5. #5
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    I've had a go, please check my answers someone!

    Top left: 38.60m
    Top right: 5.65m

    Bottom left: 171.334m
    Bottom right: 45.583m
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  6. #6
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    Quote Originally Posted by tristan1
    I've had a go, please check my answers someone!

    Top left: 38.60m
    Top right: 5.65m

    Bottom left: 171.334m
    Bottom right: 45.583m
    Hello,

    you did a very fine job! The only thing I've to apologize is I've made a mistake in my first reply.

    The equation for the bottom right question is:
    \cos 71^\circ = \frac{43.1}{x} \ \Rightarrow \ x=\frac{43.1}{\cos 71^\circ} \approx 132.38

    Best wishes to you

    Earboth
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