Question

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is ${60}^{\circ }$. If the area of the quadrilateral is $4\sqrt{3}$, find the remaining two sides

Answer 1

Let $PQRS$ be the cyclic quadrilateral with side $a=5$ and $b=2$
Also $\angle P={60}^0$ (given) and $\angle R={120}^0$ (cyclic quadrilaterl)
${SQ}^2=a^2+b^2-2ab\cos {60}^0=c^2+d^2-2cd\cos {120}^0$
Substituting the value of a and b
$c^2+d^2=19-cd$ ...(1)
Area $PQRS=4\sqrt{3}$ (given)
=Area of $△PSQ$ + area of $△QSR$
$=\frac{1}{2}ab\sin {60}^0+\frac{1}{2}cd\sin {120}^0$
$=\frac{1}{2}\left(5\right)\left(2\right)\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{2}cd\sin {120}^0$
$\Rightarrow cd=6$
$c=3,b=2$ or $c=2,d=3$
Therefore renaming two sides are 2 and 3