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Write the correct answer in each of the following:
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be

(A) $50^{\circ}$

(B) $65^{\circ}$

(C) $145^{\circ}$

(D) $155^{\circ}$

Answers (1)

best_answer

Answer: (D) $155^{\circ}$

Solution.

In $\triangle ABC$

$\angle A +\angle B + \angle C = 180^{\circ}$

[Sum of all interiors angles of triangle is $180^{\circ}$ ]

$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B + \frac{1}{2}\angle C =$ $\frac{180^{\circ}}{2}$ $=90^{\circ}$

$\Rightarrow \frac{1}{2} \angle B+ \frac{1}{2}\angle C=90^{\circ}-\frac{1}{2}\angle A................(1)$

$\text{Since, in}\triangle BOC$

$\text{As BO and OC are the angle bisectors of }\angle ABC$ $\text{and }\angle BCA$ ,

$\frac{\angle B}{2}+ \frac{\angle C}{2}+\angle BOC=180^{\circ}..................(2)$

Put value of equation (1) in equation (2)

$90^{\circ} -\frac{1}{2}\angle A+\angle BOC =180^{\circ}$
$\angle BOC = 180^{\circ} - 90^{\circ} + \frac{1}{2}\angle A$

$\angle BOC = 90^{\circ} + \frac{1}{2}\angle A$

$\therefore \angle A=130^{\circ}$

$\angle BOC= 90^{\circ}+\frac{1}{2}\times 130^{\circ}$

$\angle BOC= 90^{\circ} + 65^{\circ}$

$\angle BOC= 155^{\circ}$

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