Question

In the given figure, find $\angle BDC$.

• A. ${60}^o$
• B. ${75}^o$
• C. ${45}^o$
• D. ${50}^o$

Answer 1

Answer: (C)

${45}^o$

Let given quadrilateral $ABCD$ is a parallelogram.

Then, $AB$ is parallel to $CD$ and $AD$ is parallel to $BC$ (Opposite sides of parallelogram are parallel)

$\Rightarrow \angle$DAB $=$ $\angle$BCD (Opposite angles of parallelogram are equal)

Since, it is given that $\angle$DAB $=$75${}^{\circ }$.

Then, $\angle$BCD $=$ 75${}^{\circ }$ ---- (1)

$\Rightarrow$ $\angle$DBC = 60${}^{\circ }$ ---- (2) (Given)

In $△$DCB,

$\angle$BDC + $\angle$BCD + $\angle$DBC $=180$${}^{\circ }$

$\angle$BDC + 75${}^{\circ }$ + 60${}^{\circ }$$=$ 180${}^{\circ }$ [From (1) and (2) ]

$\angle$BDC $=$ 180${}^{\circ }$ - 135${}^{\circ }$

$\therefore$ $\angle BDC={45}^{\circ }$