Gujarat BoardEnglish MediumSTD 9MathsCircles4 Marks
Question
In a cyclic quadrilateral $ABCD$, if $\angle\text{A}-\angle\text{C}=60^\circ,$ prove that the smaller of two is $60^\circ$
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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 $(2)$
$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$
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