Question

$\sf \scriptsize{ABCD \: is \: a \: cyclic \: quadrilateral \: , in \: which \: \angle \: DBC \: = 80° \: , } \\ \sf \: \scriptsize{ \angle \: BAC \: = 40 °. \: Find \: \ \angle \: BCD. }$
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Answer 1

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$\huge\boxed{\bf{\red{\fcolorbox{blue}{white}{Answer}}}}$

$\sf \scriptsize \red{Given = }{ In \: a \: cyclic \: quadrilateral \: ABCD , \angle DBC \: = 80° , \angle \: BAC \: = 40 °}$

$\sf \scriptsize \red{To \: find \: = }{ \angle \: BCD = ?}$

$\sf \scriptsize \red{Procedure = }{ \angle \: DBC \: = \angle \: DAC \: \: \: \: \: (Angles \: in \: the \: same \: segments \: )}$

$\sf \scriptsize{ \implies \angle \: DAC = 80° … (∵ DBC = 80° )} \\ \sf \scriptsize{ \implies Now, \: \angle \:DAB = \angle \: DAC + \angle \: CAB = 80° + 40 \degree = 120 \degree} \\ \sf \scriptsize{ Then, \: \angle \: DAB + \angle \: BCD = 180 \degree....(opposite \: angles \: of \: a \: cyclic \: quadrilateral.)} \\ \sf \scriptsize{ \implies \: 120 \degree + \angle \: BCD = 180 \degree} \\ \sf \scriptsize{ \therefore \: \angle \:BCD = 180 \degree \: - 120 \degree \: = } \red{ \: 60 \degree . }$

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$\sf \scriptsize \: \red{Note = }{ Scroll \: Left \: to \: right \: side \: to \: See \: } \\ \sf \scriptsize{whole \: solution \: of \: Your \: question.}$

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