Question

In figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\angle BEC={125}^o$ and $\angle ECD={30}^o$. Then $\angle BAC=$

• A. ${95}^o$
• B. ${110}^o$
• C. ${85}^o$
• D. ${105}^o$

Answer 1

The correct option is A ${95}^o$

Since BD is a line

$\angle BEC+\angle DEC={180}^o$ (Linear pair)

${125}^o+\angle DEC={180}^o$

$\angle DEC={180}^o-{125}^o$

$\angle DEC={55}^o$

In $△DEC$

$\angle DEC+\angle EDC+\angle ECD={180}^o$ (Angle sum property of triangle)

${55}^o+\angle EDC+{30}^o={180}^o$

$\angle EDC+={180}^o-{85}^o$

$\angle EDC+={95}^o$

For segment BADCB

$\angle BAC$ & $\angle BDC$ are in the same segment

So they must be equal

$\therefore \angle BAC=\angle BDC$ (Angle in the same segment are equal)

$\therefore \angle BAC={95}^o$