Question
In the given figure, ABCD is a parallelogram in which $\angle\text{BAD}=75{^\circ}$ and $\angle\text{CBD}=60^{\circ}.$ Then, $\angle\text{BDC}=?$
- 60°
- 75°
- 45°
- 50°
✓
Answer
- 45°
We know that, the opposite angles of a parallelogram are equal.
$\therefore\angle\text{C}=\angle\text{A}=75^{\circ}$
In $\triangle\text{BCD},$
$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow75^{\circ}+\angle\text{BDC}+60^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{BDC}=45^{\circ}$
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