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Distance of Window Problem

Suppose a girl sees a 2m high window 1m above the horizontal a distance  
\[Dm\]
  away. If the girl measures the angle between the top and bottom of the window to be 30o, what is  
\[D\]
?

distance of window of known height subtending known angle

The angle subtended by the bottom 1m of wall is  
\[\alpha\]
  and  
\[tan \alpha = \frac{1}{D} \rightarrow D=\frac{1}{tan \alpha}\]
.
The angle subtended by the bottom 1m of wall and the window is  
\[\alpha + 30\]
  and  
\[tan ( \alpha +30) = \frac{3}{D} \rightarrow D=\frac{3}{tan ( \alpha + 30)}\]
.
Hence  
\[\frac{1}{tan \alpha} = \frac{3}{tan ( \alpha + 30)} \rightarrow 3tan \alpha =tan (\alpha + 30)=\frac{tan \alpha + tan 30}{1-tan \alpha tan 30}= \frac{tan \alpha + 1/\sqrt{3}}{1-tan \alpha / \sqrt{3}}\]

Multiplying top and bottom by  
\[\sqrt{3}\]
  gives  
\[3tan \alpha = \frac{\sqrt{3} tan \alpha +1}{\sqrt{3} - tan \alpha}\]
.
Now multiply both sides by  
\[\sqrt{3} - tan \alpha\]
  to give  
\[3tan \alpha (\sqrt{3} - tan \alpha=\sqrt{3} tan \alpha +1\]
.
Rearrangement gives  
\[3 tan^2 \alpha - 2 \sqrt{3} tan \alpha +1 =0 \rightarrow (\sqrt{3} tan \alpha -1)^2 =0 \rightarrow \alpha = tan^{-1}(1/ \sqrt{3}) = 30^o\]
.

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