Question

In the given figure, find
$\angle CBF$

Answer 1

$\begin{array}{c}Sumofangleof\Delta DEF={180}^0\\ \therefore \angle D+\angle DEF+\angle DFE={180}^0\\ {90}^0+{45}^0+\angle DFE={180}^0\\ \angle DFE={180}^o-{90}^0\\ ={180}^0-{135}^0={45}^0\\ DFCisastraightline\left(linearpairs\right)\\ \angle DFE+{70}^0+\angle CFB={180}^0\\ {45}^o+{70}^0+\angle CFB={180}^0\\ \angle CFB={180}^0-{45}^0-{70}^0={65}^0\\ sumofangleof\Delta CFB={180}^0\\ \angle CFB+\angle C+\angle CBF={180}^0\\ ={65}^0+{90}^0+\angle CBF={180}^0\\ \angle CBF={180}^0-{90}^0-{65}^0\\ ={25}^0\end{array}$