In $△$ ABC, bisectors of A and B intersect at point O. If $∠ C = 70^{\circ}$ . Find measure of $∠ A O B$ .

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Solution

$Given C = 70^{\circ}$
$O A B = \frac{1}{2} CAB \dots\dots(i) (AO is the bisector of CAB)$
$O B A = \frac{1}{2} CBA \dots\dots(ii) (BO is the bisector of CBA)$
$In A B C$
$C A B + C B A + C = 180^{\circ} (Angle sum property of triangle)$
$C A B + C B A + 70^{\circ} = 180^{\circ}$
$C A B + C B A = 180^{\circ} - 70^{\circ}$
$C A B + C B A = 110^{\circ}$
$Multiply both sides by \frac{1}{2}$
$\frac{1}{2} C A B + \frac{1}{2} C B A = 55^{\circ}$
$O A B + O B A = 55^{\circ} \dots(iii) (From (i) and (ii))$
$In A O B$
$A O B + O A B + O B A = 180^{\circ} (Angle sum property of triangle)$
$A O B + 55^{\circ} = 180^{\circ} (From (iii))$
$A O B = 180^{\circ} - 55^{\circ} = 125^{\circ}$
$Hence measure of AOB is 125^{\circ} .$

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In $Δ A B C,$ if bisectors of $∠ A B C$ and $∠ A C B$ intersect at O at angle of $120^{\circ}$ , then find the measure of $∠ A .$

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