In the following figure, a tower AB is $20\ m$ high and BC, its shadow on the ground, is $20\sqrt{3} \ m$ long. Find the Sun's altitude.br
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Given: Height of the tower, $AB=20\ m$ and its shadow on the ground, BC=20$\sqrt{3}$m.
To do: To find the sun's altitude.
Solution:
Let AB be the tower and BC be its shadow and $\theta $ be the sun's altitude.
$AB = 20,\ BC = 20$, $\theta=?$
In $\vartriangle ABC$,
$tan\theta=\frac{AB}{BC} \ \ \ \ \ \ \ \ \ \ \ \ ( \because tan\theta =\frac{perp.}{base})$
$\Rightarrow tan\theta =\frac{20}{20\sqrt{3}}$
$\Rightarrow tan\theta =\frac{1}{\sqrt{3}}$
But as known $tan30^{o}=\frac{1}{\sqrt{3}}$
$\Rightarrow tan\theta =tan30^{o}$
$\Rightarrow \theta =30^{o}$
The sun's altitude is at $30^{o}$.
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