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Question

In the given figure, AB || CD, ∠ABE = 120°, ∠ECD = 100° and ∠BEC = x° Find the value of x.

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Solution

$\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD}\phantom{\rule{0ex}{0ex}}\angle \mathrm{ABE}=120°\phantom{\rule{0ex}{0ex}}\angle \mathrm{ECD}=100°\phantom{\rule{0ex}{0ex}}\angle \mathrm{BEC}=\mathrm{x}°\phantom{\rule{0ex}{0ex}}\mathrm{Construction}:\mathrm{FEG}\parallel \mathrm{AB}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{sin}ceAB\parallel FEG\mathrm{and}AB\parallel CD,FEG\parallel CD\phantom{\rule{0ex}{0ex}}\therefore EFG\parallel AB\parallel CD\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{ABE}=\angle \mathrm{BEG}=120°\left(\mathrm{alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{or}\mathrm{x}°+\mathrm{y}°=120°....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{DCE}=\angle \mathrm{CEF}=100°\left(\mathrm{alternate}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{or}x°+\mathrm{z}°=100°.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{x}°+\mathrm{y}°+\mathrm{z}°=180°\left(\mathrm{FEG}\mathrm{is}\mathrm{a}s\mathrm{traight}\mathrm{line}\right)...\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right):\phantom{\rule{0ex}{0ex}}2\mathrm{x}°+\mathrm{y}°+\mathrm{z}°=220°\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{x}°+180°=220°\left(\mathrm{substi}\mathrm{tuting}\left(\mathrm{iii}\right)\right)\phantom{\rule{0ex}{0ex}}\mathrm{x}°=40°\phantom{\rule{0ex}{0ex}}\therefore x=40$

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