Question

ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle with centre O through A, B, C, D. If $\angle ADC={130}^{\circ }$, Calculate $\angle$BAC.

Answer 1

$\angle ADC+\angle ABC={180}^{\circ }$ (opposite angles are supplementary in a cyclic quadrilateral)

${130}^{\circ }$+ $\angle ABC={180}^{\circ }$

$\angle ABC={180}^{\circ }-{130}^{\circ }$

= ${50}^{\circ }$

$\angle ACB={90}^{\circ }$ (Angle in a semi circle)

In $\Delta ABC$

$\angle ABC+\angle BCA+\angle BAC={180}^{\circ }$

${50}^{\circ }$+${90}^{\circ }$+$\angle BAC={180}^{\circ }$

$\angle$ BAC = ${180}^{\circ }$-${140}^{\circ }$

= ${40}^{\circ }$