MCQ
In Fig. if $AB || CO,$ $\angle \text{CAB}=49^\circ, \angle \text{CBD}=27^\circ$ and $\angle \text{BDC}=112^\circ,$ then the values of $x$ and $y$ are:
- ✓$x = 41, y = 90$
- B$x = 41, y = 63$
- C$x = 63, y = 41$
- D$x = 90, y = 41$
✓
Answer
Correct option: A.
$x = 41, y = 90$
Since, $AB || CD$
$\angle \text{ABD}+\angle \text{CDB}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow \text{x}^\circ+27^\circ+112^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=41^\circ$
$\Rightarrow \text{x}=41$
Now, In $\triangle \text{ABC},$
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 49^\circ+41^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=90^\circ$
$\Rightarrow \text{y}=90$
Hence, the correct answer is option $(a).$
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