Question

If CD = 10 m, the length of AB is ______.

Options

• \[\frac{40}{\sqrt{3}} \mathrm{~m}\]

• \[\frac{20}{\sqrt{3}} \mathrm{~m}\]

• 100 m

• 30 m

Answer 1

If CD = 10 m, the length of AB is \[{\underline{\boldsymbol{\frac{40}{\sqrt{3}} \mathrm{~m}}}}\].

Explanation:

We know that,

\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]

From figure,

In \[\triangle BCD,\]

\[\Rightarrow \tan 30^\circ = \frac{BD}{CD}\]

\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{BD}{10}\]

\[\Rightarrow BD = \frac{10}{\sqrt{3}}\ \text{m}.\]

In \[\triangle ACD,\]

\[\Rightarrow \tan 60^\circ = \frac{AD}{CD}\]

\[\Rightarrow \sqrt{3} = \frac{AD}{10}\]

\[\Rightarrow AD = 10\sqrt{3}\ \text{cm}.\]

From figure,

\[AB = AD + BD = 10\sqrt{3} + \frac{10}{\sqrt{3}}\]

\[{} = \frac{10\sqrt{3} \times \sqrt{3} + 10}{\sqrt{3}} = \frac{30 + 10}{\sqrt{3}} = \frac{40}{\sqrt{3}}\ \text{m}.\]