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Thread: Check my work please!

  1. #1
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    Check my work please!

    The elevation of an aircraft is noted simultaneously by three observers A, B, C stationed in a straight horizontal line with AB=BC=200m. if the angles of elevation of the aircraft from A, B, and C are $\displaystyle 35^o$, $\displaystyle 40^o$, and $\displaystyle 45^o$ respectively, find the height of the aircraft.

    My work:
    $\displaystyle AQ=\frac{h}{\tan 35^o}$
    $\displaystyle BQ=\frac{h}{\tan 40^o}$
    $\displaystyle CQ=h$
    Using cosine rule,
    $\displaystyle \cos x=\frac{200^2+(\frac{h}{\tan 40^o})^2-(\frac{h}{\tan 35^o})^2}{2\times 200\times \frac{h}{\tan 40^o}}$
    $\displaystyle \cos (180^o-x)=-\cos x=\frac{200^2+(\frac{h}{\tan 40^o})^2-h^2}{2\times 200\times \frac{h}{\tan 40^o}}$
    $\displaystyle \frac{200^2+(\frac{h}{\tan 40^o})^2-(\frac{h}{\tan 35^o})^2}{2\times 200\times \frac{h}{\tan 40^o}}=-\frac{200^2+(\frac{h}{\tan 40^o})^2-h^2}{2\times 200\times \frac{h}{\tan 40^o}}$
    $\displaystyle 200^2+(\frac{h}{\tan 40^o})^2-(\frac{h}{\tan 35^o})^2=h^2-200^2-(\frac{h}{\tan 40^o})^2$
    $\displaystyle 0.192h^2=80000$
    $\displaystyle h=637.3 m$
    answer is supposed to be 934.3m
    Thanks
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  2. #2
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    forgot to add the figure attachment.
    here it is
    Attached Thumbnails Attached Thumbnails Check my work please!-18k4.bmp  
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