In the following image, #m/_CBA=54^@# and #m/_DEC=46^@#, then find #m/_DEC#?

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Answer 1 · 2017-07-26T05:36:21

#m/_DCE=80^@#

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As #AB#||#CD#, we have

#m/_CBA=m/_BCD=54^@# - alternate interior angles, #BC# being tranverse

Similarly as #BC#||#DE#, #m/_BCD=m/_CDE=54^@# - alternate interior angles, #CD# being tranverse

and as #AB#||#CD#, we have

#m/_BCA=m/_DEC=46^@# - corresponding angles, #ACE# being tranverse

Now in #DeltaCDE#, #m/_CDE=54^@# and #m/_DEC=46^@#

Hence #m/_DCE=180^@-54^@-46^@=80^@#