Question
In a cyclic quadrilateral ABCD, if $(\angle\text{B}-\angle\text{D})=60^\circ,$ show that the smaller of the two is 60°
✓
Answer
ABCD is a cyclic quadrilateral$\angle\text{B}-\angle\text{D}=60^\circ\dots(\text{i})$
And$\angle\text{B}+\angle\text{D}=180^\circ\dots(\text{ii})$
Adding (i) and (ii) we get,$2\angle\text{B}=240^\circ$
$\therefore\ \angle\text{B}=\frac{240}{2}=120^\circ$
Substituting the value of $\angle\text{B}=120^\circ$ in (i) we get$120^\circ-\angle\text{D}=60^\circ$
$\Rightarrow\ \angle\text{D}=120^\circ-60^\circ=60^\circ$
The smaller of the two angles i.e. $\angle\text{D}=60^\circ.$
And$\angle\text{B}+\angle\text{D}=180^\circ\dots(\text{ii})$
Adding (i) and (ii) we get,$2\angle\text{B}=240^\circ$
$\therefore\ \angle\text{B}=\frac{240}{2}=120^\circ$
Substituting the value of $\angle\text{B}=120^\circ$ in (i) we get$120^\circ-\angle\text{D}=60^\circ$
$\Rightarrow\ \angle\text{D}=120^\circ-60^\circ=60^\circ$
The smaller of the two angles i.e. $\angle\text{D}=60^\circ.$
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