Question Number 37906 by mondodotto@gmail.com last updated on 19/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
$${join}\:{oB}\:{so}\:<{OBC}=\mathrm{90}^{{o}} \\ $$$$<{BEO}=<{OBE}={x}\left\{{since}\:{OB}={OE}={radius}\right\} \\ $$$${x}+\mathrm{18}^{{o}} +\mathrm{90}^{{o}} +{x}=\mathrm{180}^{{o}} \\ $$$$\mathrm{2}{x}=\mathrm{72}^{{o}} \:\:\:{x}=\mathrm{36}^{{o}} \\ $$$$<{CEB}=\mathrm{36}^{{o}\:\:} <{EBC}=\mathrm{90}^{{o}} +\mathrm{36}^{{o}} =\mathrm{126}^{{o}} \\ $$