Question
✓
Answer
- 22
Solution:
$\text{AB} ⊥ \text{BC}$
$⇒\ \angle\text{ABC}=90^\circ$
$\angle\text{CAB}=32^\circ$ (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°
$⇒ \angle\text{DAB} + \angle\text{ABD} + \angle\text{BDA} = 180^\circ$
$⇒\text{x}^\circ + ^\circ32°+90^\circ + \text{x}^\circ+14^\circ = 180^\circ$
$⇒\ 2\text{x}^\circ = 180^\circ - 136^\circ$
$⇒\ 2\text{x}^\circ = 44$
$⇒\ \text{x}^\circ= 22$
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