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Question

In the given figure, AB || CD, ∠ECD = 100°, ∠EAB = 50° and ∠AEC = x°. Find the value of x. Figure

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Solution

Draw $EF\parallel AB\parallel CD$. Now, $EF\parallel AB\text{and}EA$ is the transversal. $\therefore \angle EAB+\angle AEF=180°\left[\text{Sum of consecutive interior angles is supplementary}\right]\phantom{\rule{0ex}{0ex}}⇒50°+\angle AEF=180°\phantom{\rule{0ex}{0ex}}⇒\angle AEF=180°-50°=130°$ Also, $EF\parallel CD\text{and}EC$ is the transversal. $\therefore \angle ECD+\angle CEF=180°\left[\text{Sum of consecutive interior angles is supplementary}\right]\phantom{\rule{0ex}{0ex}}⇒100°+\angle CEF=180°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=80°$ Now, $\angle AEC+\angle CEF=\angle AEF\phantom{\rule{0ex}{0ex}}⇒x°+80°=130°\phantom{\rule{0ex}{0ex}}⇒x°=130°-80°=50°\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x=50$

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