Trigonometry! #68

Denote $AB$ , $BC$ , $CD$ and $AD$ by $a$ , $b$ , $c$ and $d$ respectively.

As the quadrilateral is cyclic, $\angle D=120^{\circ}$ . Now, as the area of the quadrilateral is given to be $4\sqrt{3}$ , we have

$\frac{1}{2}ab\sin B+\frac{1}{2}cd\sin D=4\sqrt{3} \\ (5)(2)\sin60^{\circ}+cd\sin120^{\circ}=8\sqrt{3} \\ 5\sqrt{3}+\frac{\sqrt{3}}{2}cd = 8\sqrt{3} \\ 10 + cd = 16 \\ cd = 6$

Now we equate $AC$ using cosine rule, in $\Delta ABC$ and $\Delta ADC$ .

$AC^{2}=a^{2}+b^{2}-2ab\cos B=c^{2}+d^{2}-2cd\cos D \\ 25+4-(20)\cos60^{\circ}=(c+d)^{2}-2cd-12\cos120^{\circ} \\ 19=(c+d)^{2}-12+6 \\ (c+d)^{2}=25 \\ \boxed{c+d=5}$

use these two equations 0.5 a b sinx=area of the and a a+b b-2 a b=c c of a triangle

let a,b,c,d are sides of cyclic quadrilateral with a diagonal length 'p' drawn such that a,b and c,d are on either side of diagonal sum of opposite angles of cyclic quadrilateral=180 degrees for traingle with sides a,b,p 0.5 2 5 sin60=5root3/2,4+25-20cos60=p p=19; similarly for other triangle with sides p,c,d 0.5 c d sin120=4root3-(5root3/2)=3root3/2 cd=6;c c+d d-2 c d cos120=p p i.e c c+d d+6=19 i.e, c c+d*d=13

cd=6;c c+d d=13 implies c,d=3,2 c+d=5