Question

$ABCD$ is a rectangle with $\angle ABD={40}^{\circ }$. Determine $\angle DBC$.

Answer 1

Here, $ABCD$ is a rectangle and $BD$ is a diagonal.

$\angle ABD={40}^o$ [ Given ]

We know that, in rectangle measure of each angle is ${90}^o$

$\therefore$ $\angle ABD+\angle DBC={90}^o$

$\Rightarrow$ ${40}^o+\angle DBC={90}^o$

$\Rightarrow$ $\angle DBC={90}^o-{40}^o$

$\therefore$ $\angle DBC={50}^o.$