A, B, C and D are four points on a circle. The chords AC and BD cut each other at E. If \(\angle BAC=42^\circ\) and \(\angle BEC=110^\circ\), what is the measure of \(\angle ACD\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 68°
Angles in the same segment of a circle are equal: \(\angle BAC\) and \(\angle BDC\) both stand on the chord BC from the same side. Then the angle sum of \(\triangle EDC\) gives the answer.
\(\angle BDC=\angle BAC=42^\circ\), so \(\angle EDC=42^\circ\)
\(\angle DEC=180^\circ-\angle BEC=180^\circ-110^\circ=70^\circ\) (B, E, D are on one straight line)
\(\angle ECD=180^\circ-42^\circ-70^\circ=68^\circ\)
E lies on AC, so \(\angle ACD=\angle ECD=68^\circ\)
