Question
Find the value of $x$ in Figure.
✓
Answer
In the given figure, $\angle\text{BAC}=80^{\circ},\angle\text{ABC}=30^{\circ},$
$\angle\text{ACE}=\text{x} \ \text{and} \ \angle\text{ECD}=90^{\circ}$ In $\triangle\text{ABC},$
we know that, exterior angle is equal to the sum of interior opposite angles.
$\therefore\angle\text{ACD}=\angle\text{CAB}+\angle\text{ABC}$
$\Rightarrow \ \angle\text{ACE}+\angle\text{ECD}=80^{\circ}+30^{\circ}$
$[\because\angle\text{ACD}=\angle\text{ACE}+\angle\text{ECD}]$
$\Rightarrow\angle\text{ACE}+90^{\circ}=110^{\circ}$
$[\because\angle\text{ECD}=90^{\circ}]$
$\Rightarrow \angle\text{ACE}=110^{\circ}-90^{\circ}=20^{\circ}$
$\angle\text{ACE}=\text{x} \ \text{and} \ \angle\text{ECD}=90^{\circ}$ In $\triangle\text{ABC},$
we know that, exterior angle is equal to the sum of interior opposite angles.
$\therefore\angle\text{ACD}=\angle\text{CAB}+\angle\text{ABC}$
$\Rightarrow \ \angle\text{ACE}+\angle\text{ECD}=80^{\circ}+30^{\circ}$
$[\because\angle\text{ACD}=\angle\text{ACE}+\angle\text{ECD}]$
$\Rightarrow\angle\text{ACE}+90^{\circ}=110^{\circ}$
$[\because\angle\text{ECD}=90^{\circ}]$
$\Rightarrow \angle\text{ACE}=110^{\circ}-90^{\circ}=20^{\circ}$
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