Question

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

Answer 1

$\therefore$ $\angle ADC=\angle ABC={105}^o$ [Opposite angles of a parallelogram are equal]

$\angle ABC+\angle CBE={180}^o$ [Linear pair]

$\angle CBE={180}^o-{105}^o={75}^o$

Hence, $\angle CBE={75}^o$