Question

Two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is ${60}^0$.If the third side is $3$, then the fourth side is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

From $△ABC$ we have
${AC}^2={AD}^2+{DC}^2-2AB.BC\cos B$
From $△ADC$ we have
${AC}^2={AD}^2+{DC}^2-2.AD.DC\cos D$
$={AD}^2+{DC}^2-2.AD.DC\cos {120}^0$
From $△ABC$ and $△ADC$ we have
$\therefore 2^2+5^2-2\times 2\times 5\cos {60}^0=x^2+3^2-2x.3\cos ({180}^0-{60}^0)$
$\Rightarrow 4+25-20\times \frac{1}{2}=x^2+9-6x(-\cos {60}^0)$
$\Rightarrow 29-10=x^2+9-6x\left(\frac{1}{2}\right)$
$\Rightarrow 19=x^2+9-3x$
$\Rightarrow x^2+3x-10=0$
$\Rightarrow (x+5)(x-2)=0$
$\Rightarrow x\ne -5,x=2$
$\therefore x=2$