Question
In the given figure, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of $x$ and $y$.
✓
Answer
We have, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow78^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-78^\circ=102^\circ$
Now, $\angle\text{BCD}+\angle\text{DCF}=180^\circ$ [Linear pair of angles]
$\Rightarrow102^\circ=\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-102^\circ=78^\circ$
Since, $DCEF$ is a cyclic quadrilateral
Then, $x + y = 180^\circ$
$\Rightarrow 78^\circ+ y = 180^\circ$
$\Rightarrow y = 180^\circ - 78^\circ = 102^\circ$
Since, $ABCD$ is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow78^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-78^\circ=102^\circ$
Now, $\angle\text{BCD}+\angle\text{DCF}=180^\circ$ [Linear pair of angles]
$\Rightarrow102^\circ=\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-102^\circ=78^\circ$
Since, $DCEF$ is a cyclic quadrilateral
Then, $x + y = 180^\circ$
$\Rightarrow 78^\circ+ y = 180^\circ$
$\Rightarrow y = 180^\circ - 78^\circ = 102^\circ$
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