Problem Statement
In a circle with center $O$, $AB$ is a diameter and $AC$ is a chord such that $\angle BAC = 30^\circ$. The tangent at $C$ intersects $AB$ produced at $D$. Find:
(i) $\angle BCD$
(ii) $\angle ADC$
Verified Solution & Marking Scheme
Angle in Semicircle
$AB$ is diameter $\implies \angle ACB = 90^\circ$. In $\triangle ABC$: $\angle ABC = 180^\circ - (90^\circ + 30^\circ) = 60^\circ$.
Alternate Segment Theorem
Angle between tangent $CD$ and chord $BC$ equals angle in alternate segment: $\angle BCD = \angle BAC = 30^\circ$.
Find ∠ADC
In $\triangle ACD$: $\angle CAD = 30^\circ$, $\angle ACD = \angle ACB + \angle BCD = 90^\circ + 30^\circ = 120^\circ$.
$\angle ADC = 180^\circ - (30^\circ + 120^\circ) = 30^\circ$