# Please tell the answer of 3rd all parts please

$Sincethecircleisdividedinto16equalpartssothetotalanglesinthecenterwillbe16.\phantom{\rule{0ex}{0ex}}Astheangleinthecenteris360sovalueofeachanglewouldbe\frac{360}{16}=\frac{90}{4}=22.5\phantom{\rule{0ex}{0ex}}Hence\angle {A}_{1}O{A}_{2}=22.5\phantom{\rule{0ex}{0ex}}\angle {A}_{1}O{A}_{3}=2\times \angle {A}_{1}O{A}_{2}=2\times 22.5=45\phantom{\rule{0ex}{0ex}}\angle {A}_{1}O{A}_{4}=3\times \angle {A}_{1}O{A}_{2}=3\times 22.5=67.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Hence\angle {A}_{1}O{A}_{12=}11\times \angle {A}_{1}O{A}_{2}\phantom{\rule{0ex}{0ex}}Hence\angle {A}_{1}O{A}_{12}is11times\angle {A}_{1}O{A}_{2}\phantom{\rule{0ex}{0ex}}$

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