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Thread: Geometric trigonometry

  1. #1
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    Unhappy Geometric trigonometry

    Can someone please help me with the following problem? I have absolutely no clue as to how to do this!
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  2. #2
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    Trigonometry

    Hello Nancy

    In question 1, D is missing, but I assume it's where the line from C meets AB.

    So, look at $\displaystyle \triangle BCD$:

    $\displaystyle DC = BC\cos x, \quad BD = BC\sin x$

    So $\displaystyle BC(BC+BD)=2DC^2$

    $\displaystyle \Rightarrow BC(BC+BC\sin x)= 2BC^2\cos^2x$

    $\displaystyle \Rightarrow BC^2(1+\sin x) = BC^2 \times 2\cos^2 x$

    $\displaystyle \Rightarrow 2\cos^2 x = 1 + \sin x$

    Now use $\displaystyle \cos^2x = 1 - \sin^2x$:

    $\displaystyle 2(1-\sin^2x) = 1 + \sin x$

    $\displaystyle \Rightarrow 2\sin^2x + \sin x -1 =0$

    $\displaystyle \Rightarrow (2\sin x -1)(\sin x +1) = 0$

    $\displaystyle \Rightarrow \sin x = \tfrac12$, since $\displaystyle \sin x = -1$ is impossible in this case.

    $\displaystyle \Rightarrow x = 30^o$


    I think you may have written down some of the other questions wrong as well. But here are a few pointers.

    2 Note that $\displaystyle \angle CAD = x$, and then express all the lengths in terms of $\displaystyle AD$. So:

    $\displaystyle AB = AD\csc x, \quad AC = AD\sec x, \quad BC = BD+DC = ...$?

    Substitute all these values into $\displaystyle 3AB+2AC=2AD + BC$, and divide through by $\displaystyle AD$.

    I think you have written 2(b) wrong. It doesn't work out as you have written it.


    3 Express $\displaystyle BD$ and $\displaystyle BC$ in terms of $\displaystyle AB$. Again, I think you haven't written the result correctly. I think it should be $\displaystyle 2\cos^4x-3\cos^2x + 1 = 0$

    The result then factorises: $\displaystyle (2\cos^2x-1)(\cos^2x-1) = 0$


    4 Should there be some more information in the diagram? Have you written the result correctly?

    Grandad
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  3. #3
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    Sir,
    Thank u sooo much for ur response! It was a great help!

    Now that you have pointed this out, it turns out that the original diagrams did have some mistakes:

    1) In question 2 (b), I don't think it is very clear on the pdf version but the sign between 3, cosx and sinx is a minus sign.
    So, the expression is tanx = 3 - cosx - sinx
    sinx + cosx - 2

    2) The homework sheet says 2cos^4 x - 3cosx+1 = 0
    But now that I look at it (thank u for ur help again) it is probably meant to be cos^2 x. Our teacher probably missed it (He has a habbit of doing that).

    3) And yes, the C was probably meant to be a D and vice-versa. But everything else seems ok (but again, there could have been a miss print)

    Thank u again!

    Nancy
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