Question

Find lengths of BD. Given AB = 60 cm and $\angle BAD={60}^{\circ }$

Answer 1

All the sides of the quadrilateral are equal. hence the quadrilateral is a rhombus.
Each side of the rhombus = 60 cm
Let $\angle A={60}^{\circ }$
Then, $\angle C=\angle A=60$ (Opposite angles)
and, $\angle B=\angle D=x$
Sum of angles of rhombus = 360
$\angle A+\angle B+\angle C+\angle D=360$
$2x=360-120$
$x={120}^{\circ }$
$\angle B=\angle D={120}^{\circ }$
In $△ABC$
$\cos B=\frac{A{B}^2+B{C}^2-A{C}^2}{2\times AB\times BC}$
$\cos 120=\frac{{60}^2+{60}^2-A{C}^2}{2\times 60\times 60}$
$\frac{-1}{2}=\frac{{60}^2+{60}^2-A{C}^2}{2\times 60\times 60}$
$-{60}^2=2\times {60}^2-A{C}^2$
$AC=60\sqrt{3}$
$AC=103.92$ cm
In $△ABD$
$\angle A={60}^{\circ }$
$\cos A=\frac{A{B}^2+A{D}^2-B{D}^2}{2\times AB\times AD}$
$\cos 60=\frac{{60}^2+{60}^2-B{D}^2}{2\times 60\times 60}$
$\frac{1}{2}\frac{2\times {60}^2-B{D}^2}{2\times 60\times 60}$
${60}^2=2\times {60}^2-B{D}^2$
$BD=60$ cm