Question

ABCD is a rectangle such that $\angle CAB={60}^{\circ }$ and $AC=aunits$.
Find the area of rectangle ABCD.

Answer 1

In $\Delta ABC$, right-angled at B

$sin{60}^{\circ }=\frac{BC}{AC}\Rightarrow \frac{\sqrt{3}}{2}=\frac{BC}{a}\Rightarrow BC=\frac{\sqrt{3}}{2}a$

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

So area of the rectangle is
$=AB\times BC=\frac{a}{2}\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}{a}^2}{4}$