Question

In a rectangle ABCD, $AB=20$ cm, $\angle$BAC = ${60}^{\circ }$, calculate side $BC$ and diagonals $AC,BD$.

Answer 1

In $△ABC$, from the above figure

$\tan {60}^{\circ }=\frac{BC}{AB}=\sqrt{3}$

$\sqrt{3}=\frac{BC}{20}$

$\therefore BC=20\sqrt{3}cm$

$\cos {60}^{\circ }=\frac{AB}{AC}=\frac{1}{2}$

$\frac{1}{2}=\frac{20}{AC}$

$\therefore AC=40cm$

In a rectangle, the two diagonals are equal.

$\therefore AC=BD=40cm$