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$ABCD$ is a quadrilateral inscribed in a circle with centre $O$. If $\angle COD=120$ degrees and $\angle BAC=30$ degrees, then the value of $\angle BCD$ (in degrees) is

1. $89$
2. $87$
3. $86$
4. $90$

Let's draw the diagram.

Now,

• $\angle COD=120^\circ, \angle BAC=30^\circ$
• $\angle DAC=\frac{\angle COD}{2}=\frac{120^\circ}{2}=60^\circ$

$\angle DAB=\angle DAC+\angle BAC$

$\Rightarrow\angle DAB=60^\circ+30^\circ=90^\circ$

The $ABCD$ is a cyclic quadrilateral, the sum of the opposite angles will be $180^\circ$.

$\Rightarrow \angle DAB+\angle BCD=180^\circ$

$\Rightarrow 90^\circ+\angle BCD=180^\circ$

$\Rightarrow \angle BCD=180^\circ-90^\circ$

$\Rightarrow \boxed{\angle BCD=90^\circ}$

Correct Answer $:\text{D}$

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