Question

ABCD is a rectangle with $\angle ABD={40}^{\circ }$. Determine $\angle DBC+\angle BAC$
[3 MARKS]

Answer 1

Here, ABCD is a rectangle and AC and BD are the diagonals. Let the diagonals intersect at P.

$\angle ABC={90}^{\circ }$
$\Rightarrow \angle ABD+\angle DBC={90}^{\circ }$
${40}^{\circ }+\angle DBC={90}^{\circ }$
$\angle DBC={90}^{\circ }-{40}^{\circ }$

$\angle DBC={50}^{\circ }$...(1)

In $△PAB$,
AP = BP [Diagonals of a rectangle bisect each other]
$\Rightarrow \angle PAB=\angle PBA$
$\Rightarrow \angle PAB={40}^{\circ }$
$\Rightarrow \angle BAC={40}^{\circ }$...(2)

Adding (1) and (2),
$\angle DBC+\angle BAC={50}^{\circ }+{40}^{\circ }$
$\Rightarrow \angle DBC+\angle BAC={90}^{\circ }$