The two adjacent sides of a cyclic quadrilateral are `2and 5` and the angle between them is `60^0dot` If the area of the quadrilateral is `4sqrt(3)` , find the remaining two sides.

Explanation

Let ABCD be the cyclic quadrilateral in which
`AB = 2 and BC = 5, angle ABC = 60^(@)`
`angle ADC = 180^(@) - 60^(@) = 120^(@)`
Area of cyclic quadrilateral ABCD
= Area of `Delta ABC + "Area of " Delta ACD`
or `4 sqrt3 = (1)/(2) AB xx BC sin 60^(@) + (1)/(2) CD xx DA sin 120^(@)`
`= (1)/(2) 2 xx 5 xx ((sqrt3)/(2)) + (1)/(2) xy ((sqrt3)/(2))`
Where `CD = x, AD = y`
`:. xy = 6`..(i)
From `Delta ABC`, we get
`AC^(2) = AB^(2) + BC^(2) - 2AB xx BC cos 60^(@)`
`= 4 + 25 - 20 ((1)/(2)) = 19`...(ii)
Also from `Delta ACD`, we get
`AC^(2) = CD^(2) + DA^(2) - 2CD DA cos 120^(@)`
`= x^(2) + y^(2) + xy = x^(2) + y^(2) + 6` [using Eq. (i)] ..(iii)
Now from Eqs. (ii) and (iii), we have
`x^(2) + y^(2) + 6 = 19 " or " x^(2) + y^(2) = 13`...(iv)
Solving Eqs. (i) and (iv), we get
`x^(4) - 13 x^(2) + 36 = 0`
or `x^(2) = 4, 9`
or `x = 2, 3 rArr y = 3, 2`
Hence, the other two sides of the cylic quadrilateral are 3 and 2