Question

Find $\sqrt{3}AB$ in the given figure.

Answer 1

Name the triangle such that, the perpendicular lines meet at C and he fourth point is D
such that, $CD=20$
$\angle ADB={30}^{\circ }$ and $\angle BDC={30}^{\circ }$
Now, in $△ADC$
$\tan ADC=\frac{AC}{CD}$
$\tan 60=\frac{AC}{CD}$
$AC=20\sqrt{3}$
In $△BDC$
$\tan BDC=\frac{BC}{CD}$
$\tan 30=\frac{BC}{CD}$
$BC=\frac{20}{\sqrt{3}}$
Thus, $AB=AC-BC$
$AB=20\sqrt{3}-\frac{20}{\sqrt{3}}$
$AB=\frac{60-20}{\sqrt{3}}$
$AB=\frac{40}{\sqrt{3}}$
$AB=23.1$ cm