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Math Help - Work out sides of scalene triangle with only one side

  1. #1
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    Work out sides of scalene triangle with only one side

    Hi,

    This problem has me stumped...

    I have a scalene triangle with one side of 10 (opposite of hypotenuse). The topmost angle is 60 degrees, the bottom left angle is 90 and the bottom right is 30.

    How do I calculate the lengths of the other side? This is for a software development project I am working on...my maths is just too rusty!
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  2. #2
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    sides of a 30-60-90 triangle (short leg, long leg, hypotenuse) are in the ratio 1:\sqrt{3}:2
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  3. #3
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    This was just one example...the angles may be different at times (there will always be one right angle) and I will always know the other angles.

    So all I have to work with is one side + all angles, which is why I was after an equation to work it out.

    Thanks anyway.
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  4. #4
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    Quote Originally Posted by AP81 View Post
    This was just one example...the angles may be different at times (there will always be one right angle) and I will always know the other angles.

    So all I have to work with is one side + all angles, which is why I was after an equation to work it out.

    Thanks anyway.
    Can you recognise the Opposite and Adjacent sides from the Point of View of a given angle? Can you recognise the Hypotenuse?

    If so you can use the formulae

    \sin{\theta^{\circ}} = \frac{Opposite}{Hypotenuse}

    \cos{\theta^{\circ}} = \frac{Adjacent}{Hypotenuse}

    \tan{\theta^{\circ}} = \frac{Opposite}{Adjacent}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Can you recognise the Opposite and Adjacent sides from the Point of View of a given angle? Can you recognise the Hypotenuse?

    If so you can use the formulae

    \sin{\theta^{\circ}} = \frac{Opposite}{Hypotenuse}

    \cos{\theta^{\circ}} = \frac{Adjacent}{Hypotenuse}

    \tan{\theta^{\circ}} = \frac{Opposite}{Adjacent}.

    I would have used Sin/Cos/Tan if I knew more than one side. All I have is the hypotenuse to work with.
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  6. #6
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    Quote Originally Posted by AP81 View Post
    I would have used Sin/Cos/Tan if I knew more than one side. All I have is the hypotenuse to work with.
    If you're given another angle besides the right angle, you can substitute it into the sin/cos/tan side of the equation, as well as your known side, and solve for the unknown side.
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