If one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles can be
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°
(b) 65°
(c) 90°
(d) 155°
Let ∆ABC be such that ∠A = 130°.
Here, BP is the bisector of ∠B and CP is the bisector of ∠C.
$\therefore \angle \mathrm{ABP}=\angle \mathrm{PBC}=\frac{1}{2} \angle \mathrm{B}$ .......(1)
Also, $\angle \mathrm{ACP}=\angle \mathrm{PCB}=\frac{1}{2} \angle \mathrm{C}$ ............(2)
In ∆ABC,
∠A + ∠B + ∠C = 180° (Angle sum property)
⇒ 130° + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° − 130° = 50°
$\Rightarrow \frac{1}{2} \angle B+\frac{1}{2} \angle C=\frac{1}{2} \times 50^{\circ}=25^{\circ}$
⇒ ∠PBC + ∠PCB = 25° .....(3) [Using (1) and (2)]
In ∆PBC,
∠PBC + ∠PCB + ∠BPC = 180° (Angle sum property)
⇒ 25° + ∠BPC = 180° [Using (3)]
⇒ ∠BPC = 180° − 25° = 155°
Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.
Hence, the correct answer is option (d).
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