Question

The measure of angle between diameter

Answer 1

$AB\: is \:the \:diameter\Rightarrow \widehat{ACB}=90°\Rightarrow \widehat{ABC}=180°-90°-30°=60° \: (1) \\ \bigtriangleup OBC \: has \: OB=OC=R\Rightarrow \bigtriangleup OBC \: is \: an \: isosceles \: triangle \: (2) \\ (1),(2)\Rightarrow \bigtriangleup OBC \: is \: an \: equilateral \: t riangle \Rightarrow \widehat{COB} =\widehat{COD} = 60° \\ CD \: tangents \: (O) \: at \: C\Rightarrow OC\perp OD \: at \: C\Rightarrow \widehat{OCD}=90° \\ \widehat{CDO}=180°-\widehat{OCD}-\widehat{COD}=180°-90°-60°=30° \\ In\: \bigtriangleup ACD,\widehat{CAD}=\widehat{CDA}=30°\\\Rightarrow \bigtriangleup ACD\: is\: an\: isosceles\: triangle\: at\: C$