MCQ
In Figure, if $\text{AB} \perp \text{BC},$ then $x =$
- A$18$
- B$25$
- ✓$22$
- D$32$
✓
Answer
Correct option: C.
$22$
$\text{AB} \perp \text{BC}$
$\Rightarrow \ \angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (Opposite angles)
Now, in $\triangle\text{ABD}$
$\angle\text{DAB }= \text{x}^\circ+32^\circ$
$\angle\text{ABD}=90^\circ$
$\angle\text{BDA }= \text{x}^\circ+14^\circ$
In a $\triangle,$ sum of all angles = 180^\circ
$\Rightarrow \angle\text{DAB} + \angle\text{ABD} + \angle\text{BDA} = 180^\circ$
$\Rightarrow \text{x}^\circ + ^\circ32^\circ +90^\circ + \text{x}^\circ+14^\circ = 180^\circ$
$\Rightarrow \ 2\text{x}^\circ = 180^\circ - 136^\circ$
$\Rightarrow \ 2\text{x}^\circ = 44$
$\Rightarrow \ \text{x}^\circ= 22$
$\Rightarrow \ \angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (Opposite angles)
Now, in $\triangle\text{ABD}$
$\angle\text{DAB }= \text{x}^\circ+32^\circ$
$\angle\text{ABD}=90^\circ$
$\angle\text{BDA }= \text{x}^\circ+14^\circ$
In a $\triangle,$ sum of all angles = 180^\circ
$\Rightarrow \angle\text{DAB} + \angle\text{ABD} + \angle\text{BDA} = 180^\circ$
$\Rightarrow \text{x}^\circ + ^\circ32^\circ +90^\circ + \text{x}^\circ+14^\circ = 180^\circ$
$\Rightarrow \ 2\text{x}^\circ = 180^\circ - 136^\circ$
$\Rightarrow \ 2\text{x}^\circ = 44$
$\Rightarrow \ \text{x}^\circ= 22$
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