Answer : (a) \(\frac{3}{4}\)
∠CED = 120° (\(\because
\) CEDA is a cyclic quad.)
⇒ ∠BED = 60°
\(\therefore\) In ∆ EDB, ∠EDB = 90°
\(\therefore\) \(\frac{BD}{BE}\) = cos 30° ⇒ \(\frac{6}{BE}\) = \(\frac{\sqrt{3}}{2}\) ⇒ BE = 4\(\sqrt{3}\) cm.
BC = BE + CE = 4\(\sqrt{3}\) + 5\(\sqrt{3}\) = 9\(\sqrt{3}\) cm
\(\because
\) AB and CB are secants of the given circle,
BD × BA = BE × EC
⇒ 6 × BA = 4\(\sqrt{3}\) × 9\(\sqrt{3}\)
⇒ BA = 18 cm.
\(\because
\) ∠ACB = 90°, ∆ ABC is a rt. ∠d ∆
⇒ AC2 = \(\sqrt{AB^2 -BC^2}\) = \(\sqrt{{18}^2 -(9\sqrt{3})^2}\)
= \(\sqrt{324 -243}\) = \(\sqrt{81}\) = 9 cm.
\(\therefore\) AD = AB – BD = 12 cm.
\(\therefore\) AC : AD = 9 : 12 = 3 : 4
= \(\frac{3}{4}\)