A quadrilateral $ABCD$ of area ${3√3\over 4}$ is inscribed in a unit circle. If $AB=1$ and the diagonal $BD=√3$, find lengths of other sides

Question

My try : AB = 1, BD = √3 , OA = OB = OD = 1, Let AD be x

The given circle of radius 1 is also circum-circle of Δ ABD

∴ R = 1 for Δ ABD

∴ a/Sin A = 2R

∴ √3/Sin A = 2R = 2

∴ Sin A = √3/2

∴ A = 60°

and hence C = 120°

Also by cosine rule on Δ ABD,

(√3)² = 1² + x² −2x cos60°

or x² −x −2 = 0

∴ x = 2 What to do next

Answer 1

Note that $\triangle{AOB}$ is equilateral, $\triangle{BOD}$ is isosceles and $\cos \angle{OBD}=\frac{\sqrt{3}}{2} \to \angle{OBD}=30° \to \angle{ABD}=90°$. $AD$ is a diameter so $AD=2$. Can you finish?

Answer 2

You got that $AD$ is a diameter of the circle.

Now, work with $\Delta BDC$.

There is also the case $\measuredangle A=120^{\circ}.$