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In the given figure, `AB||CD, angleA=90^(@)` and `angleAEC=40^(@)`. Find `angleECD`. |
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Answer» Through C draw `CF||AE`, cutting AB at G. Now, `CF||AE` and EC is the transversal. `:. angleECG=angleAEC=40^(@)` [alternate int. `angles`]. Now, `CF||AE` and AGB is a trasversal. `:. angleFGB=angleEAG=90^(@)` [corresponding `angles`] Again, `AB||CD` and GC is the transversal. `:. angleGCD=angleFGB=90^(@)` [corresponding `angles`] `:. angleECD=angleECG+angleGCD=(40^(@)+90^(@))=130^(@)`. Hence, `angleECD=130^(@)`. |
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