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प्रश्न
Using the information given in the following figure, the measure of AE is:

विकल्प
20 cm
10 cm
\[6 \frac{2}{3} \mathrm{~cm}\]
\[20 \sqrt{3} \mathrm{~cm}\]
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उत्तर
\[\boldsymbol{6 \frac{2}{3} \mathrm{~cm}}\]
Explanation:
We know that,
\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]
From figure,
In \[\triangle ABC,\]
\[\Rightarrow \tan 60^\circ = \frac{AB}{BC}\]
\[\Rightarrow \sqrt{3} = \frac{20}{BC}\]
\[\Rightarrow BC = \frac{20}{\sqrt{3}}\ \text{cm}.\]
In rectangle BCDE,
Opposite sides of rectangle are equal.
\[\therefore DE = BC = \frac{20}{\sqrt{3}}\ \text{cm}\]
From figure,
In \[\triangle AED,\]
\[\Rightarrow \tan 30^\circ = \frac{AE}{DE}\]
\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{AE}{\frac{20}{\sqrt{3}}}\]
\[\Rightarrow AE = \frac{1}{\sqrt{3}} \times \frac{20}{\sqrt{3}} = \frac{20}{3} = 6\frac{2}{3}\ \text{cm}.\]
