Question

$\sf \scriptsize{ABCD \: is \: a \: cyclic \: quadrilateral \: whose \: side \: AB \: is \: a \: diameter \: of \: the \: circle \: through \: A ,B,C, \: D. } \\ \sf \scriptsize{If \: angle \: \angle ADC = 130 \: Find \: angle \: \: \angle \: BAC.}$
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Answer 1

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

$\sf \scriptsize \red{Given = }{ ABCD \: is \: a \: cyclic \: quadrilateral \: whose \: side \: AB \: is \: a \: diameter.} \\ \sf \scriptsize { \angle \: ADC = 130 \degree }$

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

$\sf \scriptsize \red{Procedure = }{ \angle \: D \: + \angle \: B = 180 ° \: \: \: \: ....(Opposite \: angles \: of \: a \: cyclic \: quadrilateral. \: )}$

$\sf \scriptsize{As \: \: \angle \: D = 130° … (given) }\\ \sf \scriptsize { \angle\:B = 180° - 130° = 50°}$

$\sf \scriptsize { \angle \: ACB = 90° \: \: \: \: \: \: \: (Angles \: in \: a \: semicircle \: )} \\ \sf \scriptsize{In \: \triangle \: ABC , \angle \: BAC + \angle \: ABC + \angle \: ACB = 180° [Sum\: of\: three\: angle\: of\: a\: \triangle \:is\: always\: 180° ]} \\ \sf \scriptsize{ \implies\angle \: BAC + 50°+90° = 180°} \\ \sf \scriptsize{ \implies \: \angle \: BAC = 180° - 90° -50° = }\red{ \: 40°.}$

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$\sf \scriptsize \red{Note = }{ Scroll \: From \: Left \: to \: right \: side \: to \: see \: the} \\ \sf \scriptsize{ \: whole \: Solution\:of \:your \: question. }$

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