Question

In the diagram below of Rhombus ABCD,the mC = 100. Find m2DBC.

Answer 1

`\begin{gathered} \text{ Since it is a Rhombus:} \\ DC=BC \\ \text{Therefore:} \\ m\angle CBD=m\angle DBC \\ \text{For the }\Delta\text{CDB:} \\ 100+m\angle CBD+m\angle DBC=180 \\ 100+m\angle DBC+m\angle DBC=180 \\ 100+2m\angle DBC=180 \\ m\angle DBC=(180-100)/(2) \\ m\angle DBC=40 \end{gathered}`