Question

The diagonals AC & BD of parallelogram ABCD intersect each other at P of $\angle DAC={40}^o$ & $\angle APB={80}^o$. Then $\angle DBC$ is equal to ___ .

• A. ${40}^o$
• B. ${80}^o$
• C. ${120}^o$
• D. None

Answer 1

Answer: (A)

${40}^o$

ABCD in parallelogram

$AD||BC\therefore \angle DAC=\angle BCA$

$\angle APB=\angle ACB+\angle DBC$ (Exterior angle of a triangle)

$\angle DBC=\angle APB-\angle ACB$

$\angle DBC={80}^o-\angle DAC$

$\angle DBC={80}^o-{40}^o$

$\angle DBC={40}^o$