Question

In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD produced to E such that $AED={95}^o$ and $\angle OBA={30}^o$. Find $\angle OAC$.

Answer 1

In the figure, ABCD is a cyclic quadrilateral CD is produced to E such that $\angle ADE={95}^o$ O is the centre of the circle

$\because \angle ADC+\angle ADE={180}^o$

$\Rightarrow \angle ADC+{95}^o={180}^o$

$\Rightarrow \angle ADC={180}^o-{95}^o={85}^o$

Now are ABC subtends $\angle AOC$ at the centre and $\angle ADC$ at the remaining part of the circle

$\therefore \angle AOC=2\angle ADC=2\times {85}^o={170}^o$

Now in $\Delta OAC$,

$\angle OAC+\angle OCA+\angle AOC={180}^o$ (Sum of angles of a triangle)

$\Rightarrow \angle OAC=\angle OCA$ $(\because OA=OC$ radii of circle )

$\therefore \angle OAC+\angle OAC+{170}^o={180}^o$

$2\angle OAC={180}^o-{170}^o={10}^o$

$\therefore \angle OAC=\frac{{10}^o}{2}=5^o$