Question
In a cyclic quadrilateral ABCD, if $\angle\text{A}-\angle\text{C}=60^\circ,$ prove that the smaller of two is $60^{\circ}$
✓
Answer
We have$\angle\text{A}-\angle\text{C}=60^\circ\dots(1)$
Since, ABCD is a cyclic quadrilateral Then $\angle\text{A}+\angle\text{C}=180^\circ\dots(2)$ Add equations (1) and (2)$\angle\text{A}-\angle\text{C}+\angle\text{A}+\angle\text{C}=60^\circ+180^\circ$
$\Rightarrow2\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=\frac{240^\circ}{2}=120^\circ$
Put value of $\angle\text{A}$ in equation (2)$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$
Since, ABCD is a cyclic quadrilateral Then $\angle\text{A}+\angle\text{C}=180^\circ\dots(2)$ Add equations (1) and (2)$\angle\text{A}-\angle\text{C}+\angle\text{A}+\angle\text{C}=60^\circ+180^\circ$
$\Rightarrow2\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=\frac{240^\circ}{2}=120^\circ$
Put value of $\angle\text{A}$ in equation (2)$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$
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