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Math Help - Finding the length of a triangle's sides ?

  1. #1
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    Finding the length of a triangle's sides ?

    Hello

    I am using the Sine rule to find the sides of a triangle and am stuck on one part.



    How do you get the values which the two red arrows are pointing towards? I understand the rest, but have no clue how those 2 values have been worked out at all.

    Really grateful for the help I have been getting here. I have a test tomorrow, so am panicking a bit.

    Many thanks!

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  2. #2
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    Quote Originally Posted by Swerve View Post
    Hello
    [snip]

    How do you get the values which the two red arrows are pointing towards? I understand the rest, but have no clue how those 2 values have been worked out at all.
    [snip]
    The two red arrows are pointing to the values of \sin 45^0 and \sin 105^0. You get these values by using the sin button on your calculator, making sure the calculator is in degree mode, and rounding to three decimal places.
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  3. #3
    Senior Member DivideBy0's Avatar
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    \sin{45}=\frac{\sqrt{2}}{2}

    \sin{105}=\sin{(60+45)}=\sin{60}\cos{45}+\cos{60}\  sin{45}=(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})+(  \frac{1}{2})(\frac{\sqrt{2}}{2})

    =\frac{\sqrt{6}+\sqrt{2}}{4}
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    \sin{45}=\frac{\sqrt{2}}{2}

    \sin{105}=\sin{(60+45)}=\sin{60}\cos{45}+\cos{60}\  sin{45}=(\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})+(  \frac{1}{2})(\frac{\sqrt{2}}{2})

    =\frac{\sqrt{6}+\sqrt{2}}{4}
    No offence to anyone but ..... DivideBy0, I think you're living in the Penthouse Suite whilst Swerve is still trying to check in at the lobby .....
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