## 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
$D$
?

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
$3 tan^2 \alpha - 2 \sqrt{3} tan \alpha +1 =0 \rightarrow (\sqrt{3} tan \alpha -1)^2 =0 \rightarrow \alpha = tan^{-1}(1/ \sqrt{3}) = 30^o$
.