Question

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°

Answer 1

Let ∆ABC be such that ∠A = 130°.



Here, BP is the bisector of ∠B and CP is the bisector of ∠C.

∴ ∠ABP = ∠PBC = $\frac{1}{2}$∠B .....(1)

Also, ∠ACP = ∠PCB = $\frac{1}{2}$∠C .....(2)

In ∆ABC,

∠A + ∠B + ∠C = 180° (Angle sum property)

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

⇒ $\frac{1}{2}$∠B + $\frac{1}{2}$∠C = $\frac{1}{2}$ × 50° = 25°

⇒ ∠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).