If one of the angles of a triangle is 130^{ci | vedclass

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(B) Let the triangle be $	riangle ABC$ with $\angle A = 130^{\circ}.$In any triangle,the sum of all interior angles is $180^{\circ},$ so $\angle B +

Vedclass · Accepted Answer

$155$

Vedclass · Answer

(B) Let the triangle be $	riangle ABC$ with $\angle A = 130^{\circ}.$In any triangle,the sum of all interior angles is $180^{\circ},$ so $\angle B + \angle C = 180^{\circ} - 130^{\circ} = 50^{\circ}.$Let $OB$ and $OC$ be the bisectors of $\angle B$ and $\angle C$ respectively.Then $\angle OBC = \frac{1}{2} \angle B$ and $\angle OCB = \frac{1}{2} \angle C.$In $	riangle OBC,$ the sum of angles is $180^{\circ},$ so $\angle BOC = 180^{\circ} - (\angle OBC + \angle OCB).$Substituting the values,$\angle BOC = 180^{\circ} - \frac{1}{2}(\angle B + \angle C).$$\angle BOC = 180^{\circ} - \frac{1}{2}(50^{\circ}) = 180^{\circ} - 25^{\circ} = 155^{\circ}.$

If one of the angles of a triangle is $130^{\circ},$ then the angle between the bisectors of the other two angles can be (in $^{\circ}$)

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