Question

In the figure, if $\angle BAC={60}^{o}and\angle BCA={20}^o,find\angle ADC$

Answer 1

In $\Delta ABC$, $\angle BAC+\angle ABC+\angle ACB={180}^o$ (Sum of angles of triangle)

$\Rightarrow {60}^o+\angle ABC+{20}^o={180}^o$

$\Rightarrow \angle ABC+{80}^o={180}^o$

$\therefore \angle ABC={180}^o-{80}^o={100}^o$

$\because$ ABCD is a cyclic quadrilateral,

$\therefore \angle ABC+\angle ADC={180}^o$

$\Rightarrow {100}^o+\angle ADC={180}^o$

$\therefore \angle ADC={180}^o-{100}^o={80}^o$