In the figure below, AB is parallel to DC, and BC is parallel to ED If mDCB = 90°, mDCE = 35° and mBAC = 49°, what is mECA?

Question

 

In the figure below, AB is parallel to DC, and BC is parallel to ED.

Note: picture not drawn to scale

If mDCB = 90°, mDCE = 35° and mBAC = 49°, what is mECA?

Answer 1

Angle DCB (=90) is made up of angles DCE+ECA+ACB. Angle ACB=90-BAC=90-49=41 degrees. So 35+ECA+41=90. Therefore angle ECA=90-35-41=14 degrees. The other way to work this out is to note that angle BAC=angle ACD because AB and DC are parallel and BAC and ACD are alternate angles=49 degrees. So angle ECA=ACD-DCE=49-35=14 degrees.

Answer 2

Given AB is parallel to DC,
and BC is parallel to ED

If mDCB = 90°,
mDCE = 35°
and mBAC = 49°

All angle of a rectangle is 360°
90°+35°+49°+x=360°
x=186°

Hence the angle,
mECA=186°

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