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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"

Academic Mathematics NCERT Class 10

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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Updated on: 10-Oct-2022

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