Suppose a girl sees a 2m high window 1m above the horizontal a distance  {jatex options:inline}Dm{/jatex}  away. If the girl measures the angle between the top and bottom of the window to be 30o, what is  {jatex options:inline}D{/jatex}?

The angle subtended by the bottom 1m of wall is  {jatex options:inline}\alpha{/jatex}  and  {jatex options:inline}tan \alpha = \frac{1}{D} \rightarrow D=\frac{1}{tan \alpha}{/jatex}.
The angle subtended by the bottom 1m of wall and the window is  {jatex options:inline}\alpha + 30{/jatex}  and  {jatex options:inline}tan ( \alpha +30) = \frac{3}{D} \rightarrow D=\frac{3}{tan ( \alpha + 30)}{/jatex}.
Hence  {jatex options:inline}\frac{1}{tan \alpha} = \frac{3}{tan ( \alpha + 30)} \rightarrow 3tan \alpha =tan (\alpha + 30)=\frac{tan \alpha + tan 30}{1-tan \alpha tan 30}= \frac{tan \alpha + 1/\sqrt{3}}{1-tan \alpha / \sqrt{3}}{/jatex}
Multiplying top and bottom by  {jatex options:inline}\sqrt{3}{/jatex}  gives  {jatex options:inline}3tan \alpha = \frac{\sqrt{3} tan \alpha +1}{\sqrt{3} - tan \alpha}{/jatex}.
Now multiply both sides by  {jatex options:inline}\sqrt{3} - tan \alpha{/jatex}  to give  {jatex options:inline}3tan \alpha (\sqrt{3} - tan \alpha=\sqrt{3} tan \alpha +1{/jatex}.
Rearrangement gives  {jatex options:inline}3 tan^2 \alpha - 2 \sqrt{3} tan \alpha +1 =0 \rightarrow (\sqrt{3} tan \alpha -1)^2 =0 \rightarrow \alpha = tan^{-1}(1/ \sqrt{3}) = 30^o{/jatex}.