Question

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60$.

If the third side is $3$, then the remaining fourth side is :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Let AB = 2 and BC = 5
$\angle ABC={60}^{\circ }$ (given).
Since the quadrilateral is cyclic,
$\angle CDA={180}^{\circ }-{60}^{\circ }={120}^{\circ },$
Let CD = c and DA = d
Also $A{B}^2+B{C}^2-2AB.BCcos{60}^{\circ }=A{C}^2$
$=C{D}^2+D{A}^2-2CD.DAcos120$
by cosine rule.
or $4+25-2.2.5.\frac{1}{2}=c^2+d^2+cd$
$19=c^2+d^2+cd=9+d^2+3d$
$\therefore d^2+3d10=0$
or $(d+5)(d-2)=0\therefore d=2$