Question

find x in the following​

Answer 1

$\huge\rm\green{answeR:}$

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$\bf{{\underline{Given}}:}$

• $\sf{OD~is~the~bisector~of~∠BOC}$

$\sf\implies{∠BOD=∠COD=x}$

$\bf{{\underline{To~find}}:}$

• $\sf{The~value~of~~‛x’}$

$\sf{~~~~~~~~~~ ∠COD+∠BOD+∠BOF=180°~~~~(Straight~line)}$

$\sf\implies{x+x+20=180}$

$\sf\implies{2x+20=180}$

$\sf\implies{2x=180-20}$

$\sf\implies{2x=160}$

$\sf\implies{x={\dfrac{160}{2}}}$

$\sf\implies{x={\green{\underline{\boxed{\bf 80°}}}}}$

$\bf\therefore{{\underline{Required~answer}}:}$

• $\sf{x={\green{\underline{\boxed{\bf 80°}}}}}$

Answer 2

${\huge{\red{\boxed{\green{\boxed{\blue{\boxed{\orange{\boxed{\pink{\boxed{\mathcal\purple{Answer}}}}}}}}}}}}}$

$Given :$

$OD \:is \:the \:bisector \:of \:∠BOC$

$⟹BOD=∠COD=x$

$To \:find :$

$The \:value \:of \:‛x’$

$∠COD+∠BOD+∠BOF=180° (Straight line)$

$⟹x+x+20=180$

$⟹2x+20=180$

$⟹2x=180−20$

$⟹2x=160$

$⟹x= 160/2$

$⟹x= 80°$