Question

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ units and the angle between them is ${60}^0$.

lf the third side is $3$ units, the remaining fourth side is?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

First draw $△ABC$ with $AB=2,BC=5and\angle ABC={60}^o.$ Then find its circum center O and then draw the circle. Then from C draw an arc of radius 3 units to cut the circle at D. Join AD. We have to find AD.

$△ABC:$ Draw $AE\perp BC.$

$AE=2\ast sin{60}^o=\sqrt{3}$

and $BE=2\ast cos{60}^o=1$

So $EC=4$

$△AEC:$ $A{C}^2=A{E}^2+E{C}^2=19,$

$AC=\sqrt{19}$

$△ADC:$ Draw $AF\perp CD.$

Let $AD=x$. $\angle ADC=120°(since,{180}^o-\angle ABC)$

$△ADF:$ $\angle ADF=60°$ ,

$So,AF=xsin{60}^o=\sqrt{3}x/2$

$FD=xcos{60}^o=x/2$

$△AFC:AF²=AC²-FC²$

$(\sqrt{3}x/2{)}^2=19-(3+x/2{)}^2$

$3x^2/4=19-9-x^2/4-3x$

$x^2+3x-10=0$

$(x+5)(x-2)=0$

Side cannot have negative length , so answer is $2$

Therefore, Correct Option is $A$