Question

# In fig. 1.54 $\angle COD$ = 90$°$ $\angle$BOE = 72$°$ and $\angle$AOB is a straight angle. Find the measures of the following angles. $\angle$AOC, $\angle$BOC, $\angle$AOE

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Solution

## Given: $\angle \mathrm{COD}=90°\phantom{\rule{0ex}{0ex}}\angle \mathrm{BOE}=72°\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOB}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{angle}.$ $\angle \mathrm{AOC}=\angle \mathrm{BOE}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{AOC}=72°$ $\mathrm{Here},\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180°\left(\angle \mathrm{AOB}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{angle}\right)\phantom{\rule{0ex}{0ex}}⇒72°+90°+\angle \mathrm{BOD}=180°\phantom{\rule{0ex}{0ex}}⇒162°+\angle \mathrm{BOD}=180°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{BOD}=18°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\angle \mathrm{BOC}=\angle \mathrm{BOD}+\angle \mathrm{COD}\phantom{\rule{0ex}{0ex}}=18°+90°\phantom{\rule{0ex}{0ex}}=108°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ $\mathrm{Thus},\angle \mathrm{AOE}=\angle \mathrm{BOC}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.,\angle \mathrm{AOE}=108°$

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