Question

in the figure ,A0=0D= 0C and reflex AOC =230 FIND THE VALUE OF X​

Answer 1

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GIVEN

$\rightarrow \angle AOC_{reflex} =230\degree\\ \rightarrow AO=OD=OC\\ \rightarrow x=?$

•FOr the ∆ AOD&∆ COD,

$\rightarrow OD=OD\:\:(\because common\: sides)\\ \rightarrow AO=OC\:\:(given)\\ and \rightarrow AO=OD=OC\:\:(given)\\ \triangle AOD &#8776 \triangle COD\\ \therefore \angle OAD= \angle OCD\:\:(c.p.c.t.)$

•Now, for ∆ AOD,

AO=OD

so, $\rightarrow \angle ADO= \angle OAD\\</p><p>$

Now, for ∆ OCD

OD =OC

so, $\rightarrow \angle ODC= \angle OCD\\ \rightarrow \angle ODC=\angle OAD</p><p>$

Now,

$\therefore \angle ADO=\angle ODC$

Now ,,,

$\rightarrow \angle AOC=360\degree-\angle AOC_{reflex}\\ \rightarrow \angle AOC=360\degree-230\degree\\ \rightarrow \angle AOC=130\degree$

therefore....

$\bf\implies \angle AOD=180\degree-(\angle OAD+\angle ODA)\\ \bf\implies \angle AOD=180\degree-(\angle OCD+\angle ODC)\\</p><p> \bf\implies \angle AOD=\angle COD\\</p><p> \bf\implies \angle AOD=\angle COD=\frac{1}{2}\angle AOC\\now \:given\:That\:\:\angle AOC= 130\degree\\ \therefore \bf\implies \angle AOD=\angle COD=\frac{1}{2}\times 130\degree\\ \bf\implies \angle AOD=\angle COD=65\degree$

Now from the ∆AOD:-

$\bf\rightarrow \angle ADO+\angle OAD+\angle AOD=180\degree \\ \bf \rightarrow \angle ADO+\angle ADO=180\degree-65\degree\\ (\because \angle ADO=\angle OAD\:\:(proved \:earlier) \\ \bf</p><p>\rightarrow 2\angle ADO=115\degree\\ \bf\rightarrow \angle ADO=\frac{1}{2}\times 115\degree\\ \bf \rightarrow \boxed{\bf\red{\angle ADO=\frac{115\degree}{2}}}$

$\therefore \rightarrow\boxed{\bf\red{\angle ODC=\frac{115\degree}{2}}} \:\:(\because \angle ADO=\angle ODC)$

$\therefore \angle ADC=\angle ADO+\angle ODC\\ \implies \angle ADC=\frac{115\degree}{2}+\frac{115\degree}{2}\\ \implies \boxed{\boxed{\green{\bf\angle ADC=115\degree}}}$

Now.......

$\therefore \implies \angle ADC+\angle ABC=180\degree\:\:\\(\because sum \:opposite \:angles\: of\: \\a\: parallelogram\: Is =180\degree ) \\ \implies 115\degree+x=180\degree\\ \implies x=180\degree-115\degree\\ \implies \boxed{\large\mathcal\pink{\bf x=65\degree}}$

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