Question

Question #e9bd2

Answer 1

`/_CDO=35^o`.

Explanation:

The diagonals [ `AC` & `BD` ] of rhombus [ `ABCD` ] intersect [ at `O` ] at Right Angles.

Now consider `DeltaAOB`, in which, `(1)/_AOB=90^o, (2) /_OAB=55^o [ Given ]`, giving `(3)/_OBA=35^o`

Notice that, `DC || AB`, & `DB` is a transvers, making congruent pair of angles `/_CDB` and `/_DBA.`

`:.` Reqd. Angle = `/_CDO`=`/_CDB`=`/_DBA`=`/_OBA=35^o,` using `(3)`.