MCQ
In the given figure, $ABCD$ is a parallelogram in which $\angle\text{BAD} = 75^\circ$ and $\angle\text{CBD} = 60^\circ.$ Then, $\angle\text{BDC} =\ ?$
- ✓$45^\circ$
- B$75^\circ$
- C$50^\circ$
- D$60^\circ$
✓
Answer
Correct option: A.
$45^\circ$
It is given in the question that,
In parallelogram $ABCD$: $\angle\text{BAD} = 75^\circ, \ \angle\text{CBD} = 60^\circ$
Now, $\angle\text{DAB} = \angle\text{DCB} = 75^\circ$ (Opposite angles)
Also, in triangle DBC we know that sum of angles of a triangle is $180^\circ $
$\angle\text{DBC} + \angle\text{BDC} + \angle\text{DCB} = 180^\circ$
$60^\circ + \angle\text{BDC} + 75^\circ= 180^\circ$
$135^\circ + \angle\text{BDC} = 180^\circ$
$\angle\text{BDC} = 180^\circ – 135^\circ$
$\angle\text{BDC} = 45^\circ$
Hence, $45^\circ $ is correct.
In parallelogram $ABCD$: $\angle\text{BAD} = 75^\circ, \ \angle\text{CBD} = 60^\circ$
Now, $\angle\text{DAB} = \angle\text{DCB} = 75^\circ$ (Opposite angles)
Also, in triangle DBC we know that sum of angles of a triangle is $180^\circ $
$\angle\text{DBC} + \angle\text{BDC} + \angle\text{DCB} = 180^\circ$
$60^\circ + \angle\text{BDC} + 75^\circ= 180^\circ$
$135^\circ + \angle\text{BDC} = 180^\circ$
$\angle\text{BDC} = 180^\circ – 135^\circ$
$\angle\text{BDC} = 45^\circ$
Hence, $45^\circ $ is correct.
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