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.$\angle\text{OAB}=30^\circ$
$\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$
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.$\angle\text{OAB}=30^\circ$
$\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$
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