Question
In Figure. find the values of $a, b$ and $c.$
✓
Answer
In figure. $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ} [$sum of all angles of a triangle is $180^\circ ]$
$\Rightarrow 90^\circ + a + 70^\circ = 180^\circ $
$[\because\angle\text{A}=90^\circ $ and $\angle\text{C }=70^\circ ,$ from the figure$]$
$\Rightarrow a + 160^\circ = 180^\circ $
$\Rightarrow 180^\circ - 160^\circ = 20^\circ $
Since, $C$ is an exterior angle of $\triangle\text{ABD}.$
$\therefore \ \angle\text{c}=\text{a}+30^{\circ}=20^{\circ}+30^{\circ}=50^{\circ}$
$[\because$ exterior angle = the sum of opposite interior angles$]$
Also, b is an exterior angle of $\triangle\text{ADC},$
$\therefore \ \angle\text{b}=60^{\circ}+70^{\circ}=130^{\circ}$
$[\because$ exterior angles = sum of opposite interior angles$]$
$\Rightarrow 90^\circ + a + 70^\circ = 180^\circ $
$[\because\angle\text{A}=90^\circ $ and $\angle\text{C }=70^\circ ,$ from the figure$]$
$\Rightarrow a + 160^\circ = 180^\circ $
$\Rightarrow 180^\circ - 160^\circ = 20^\circ $
Since, $C$ is an exterior angle of $\triangle\text{ABD}.$
$\therefore \ \angle\text{c}=\text{a}+30^{\circ}=20^{\circ}+30^{\circ}=50^{\circ}$
$[\because$ exterior angle = the sum of opposite interior angles$]$
Also, b is an exterior angle of $\triangle\text{ADC},$
$\therefore \ \angle\text{b}=60^{\circ}+70^{\circ}=130^{\circ}$
$[\because$ exterior angles = sum of opposite interior angles$]$
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