In the given figure, ABCD is a quadrilateral inscribed in a circl…

Question

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

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Answer

We are given $ABCD$ is a quadrilateral with center $O$, $\angle\text{ADE}=95^\circ$ and $\angle\text{OBA}=30^\circ$
We need to find $\angle\text{OAC}$
We are given the following figure:

Since $\angle\text{ADE}=95^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-95^\circ=85^\circ$
Since squo; $ABCD$ is cyclic quadrilateral
This means
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABO}+\angle\text{OBC}+\angle\text{ADC}=180^\circ$
$\Rightarrow30^\circ+\angle\text{OBC}+85^\circ=180^\circ$
$\Rightarrow\angle\text{OBC}=180^\circ-115^\circ=65^\circ$
Since $OB = OC$ (radius)
$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=65^\circ$
In $\triangle\text{OBC}$
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\times65^\circ=180^\circ$
$\angle\text{BOC}=180^\circ-130^\circ$
$\angle\text{BOC}=50^\circ$
Since $DBAC$ and $DBOC$ are formed on the same base which is chord.
So,
$\angle\text{BAC}=\frac{\angle\text{BOC}}{2}$
$=\frac{50^\circ}{2}$
$\angle\text{BAC}=25^\circ$
Consider $\triangle\text{BOA}$ which is isosceles triangle.
$\angle\text{OAB}=30^\circ$
$\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$