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Thread: Trig problem I'm stuck on

  1. #1
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    Trig problem I'm stuck on

    I have a triangle that has internal angles of 35, 80 and 65 degrees. The area of the triangle needs to be 4000sq metres. Say I need to fence this triangle. Will 285 metres of fencing material be enough? If this is not enough fencing material can I change the side lengths of the triangle such that the area is still 4000sq m and no more than 285m of material is used? I'm not able to change the angles in this triangle by more than +or- 25 degrees.
    Even just a suggestion as to where to start tackling this problem would be greatly appreciated.
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  2. #2
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    Hello Eleven Eleven
    Quote Originally Posted by Eleven Eleven View Post
    I have a triangle that has internal angles of 35, 80 and 65 degrees. The area of the triangle needs to be 4000sq metres. Say I need to fence this triangle. Will 285 metres of fencing material be enough? If this is not enough fencing material can I change the side lengths of the triangle such that the area is still 4000sq m and no more than 285m of material is used? I'm not able to change the angles in this triangle by more than +or- 25 degrees.
    Even just a suggestion as to where to start tackling this problem would be greatly appreciated.
    I don't think you can quite do this with 285 m - the mimimum length of fencing you need will be 288 m.

    Here's the calculation.

    Suppose the triangle is $\displaystyle ABC$ with $\displaystyle A = 35, B = 80, C = 65$.

    The area of the triangle is $\displaystyle \Delta= \tfrac12ab\sin C$, and using the Sine Rule $\displaystyle b = \frac{a\sin B}{\sin A}$. So
    $\displaystyle \Delta = \frac12\frac{a^2\sin B \sin C}{\sin A}$

    $\displaystyle \Rightarrow a^2 = \frac{2\Delta\sin A}{\sin B \sin C}$

    $\displaystyle =\frac{8000 \sin 35}{\sin 80\sin65}$

    $\displaystyle \Rightarrow a = 71.70$
    This gives $\displaystyle b = \frac{a\sin B}{\sin A}=123.11$ and $\displaystyle c=\frac{a\sin C}{\sin A}=113.29$. The perimeter is then $\displaystyle 308.1$ m.

    The minimum perimeter for a given area will be when the triangle is equilateral, all angles then being $\displaystyle 60^o$. But this gives each side as $\displaystyle 96.11$ m, with a perimeter of just over $\displaystyle 288$ m.

    Grandad
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  3. #3
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    Thanks

    That's great. It's got me started so thank you!
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