Question

In the given figure, ABCD is a parallelogram in which $\angle DAB={70}^o$ and $\angle DBC={65}^o$, then find the measure of $\angle CDB$.

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

The correct option is B ${45}^o$

$\angle A+\angle B={180}^o$

${70}^o+\angle B={180}^o$

$\angle B={110}^o$

$\angle ABD+\angle DBC={110}^o$

$\angle ABD+{65}^o={110}^o$

$\angle ABD={45}^o$

Now, $\angle CDB=\angle ABD={45}^o$