MCQ
In a Quadrilateral $ABCD$, $\angle\text{A} = \angle\text{C},\ \angle\text{B} = 2\angle\text{A}, \ \angle\text{D} = \frac{1}{21}\angle\text{A}.$ Then $\angle\text{A}, \ \angle\text{B}, \ \angle\text{C} $ and $\angle\text{D}$ respectively are:
  • $80^\circ,\ 160^\circ,\ 80^\circ,\ \text{and}\ 40^\circ$
  • B
    $20^\circ,\ 30^\circ,\ 160^\circ,\ \text{and}\ 160^\circ$
  • C
    $100^\circ,\ 100^\circ,\ 80^\circ,\ \text{and}\ 80^\circ$
  • D
    $90^\circ,\ 90^\circ,\ 90^\circ,\ \text{and}\ 90^\circ$

Answer

Correct option: A.
$80^\circ,\ 160^\circ,\ 80^\circ,\ \text{and}\ 40^\circ$

$\angle\text{A} + \angle\text{B} + \angle\text{C} + \angle\text{D} = 360^\circ$ (angle sum property)
$\angle\text{A} + 2\angle\text{A} + \angle\text{A} + \frac{1}{2}$ of $\angle\text{A} = 360^\circ$
$\frac{9}{2}$ of $\angle\text{A} = 360^\circ$
$\angle\text{A} = 80^\circ$
So, $\angle\text{B}=2(80^\circ)=160^\circ,\ \angle\text{C}=80^\circ$ and $\angle\text{D} = \frac{1}{2}$ of $80^\circ=40^\circ$

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