1. ## 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

2. forgot to add the figure attachment.
here it is