Question

In the given figure, ADEC is a cyclic quadrilateral. CE and AD are extended to meet at B. ∠CAD = 60° and ∠CBA = 30°. BD = 6 cm and CE = 5 \(\sqrt{3}\) cm. What is the ratio AC : AD?

(a) \(\frac{3}{4}\) 

(b) \(\frac{4}{5}\) 

(c) \(\frac{2\sqrt{3}}{5}\)

(d) cannot be determined

Answer 1

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 

= \(\frac{3}{4}\)