Question

In the figure, ABCD is a parallelogram in which $\angle ADC={105}^{\circ }$. The value of $\angle CBE$ is

Answer 1

ABCD is a parallelogram
$\therefore ,\angle ADC=\angle ABC={105}^{\circ }$[Opposite angles of a parallelogram are equal]
$\angle ABC+\angle CBE={180}^{\circ }$ [Linear pair]
$\angle CBE={180}^{\circ }-{105}^{\circ }={75}^{\circ }$
Hence, $\angle CBE={75}^{\circ }$