Question

30. In ∆ABC, internal bisectors of angle B and C are y=x-1
and x-1 = 0, respectively. The angle A of ∆ABC is
(a) 90°
(b) 60°
(c) 120°
(d) None of these​

Answer 1

Answer:

From △BOC, since sum of three angles of a triangle is 180°.

∠BOC=180° −∠OBC−∠OCB

Since, AX, BY and CZ are internal bisector of the angles ∠A,∠B and ∠C.

∠BOC=180°−(∠B/2+∠C/2)

∠BOC=180°−(180°−∠A)/2

Given ∠A=50° ⟹ ∠BOC=115°

Now, vertically opposite angles are equal to each other.

∠BOC=∠ZOY=105°

From △ZOY,

∠OZY+∠ZOY+∠OYZ=180°

Since, ∠OZY and ∠CZY are same angle.

∠OYZ and ∠BYZ are same angle.

∠CZY+∠ZOY+∠BYZ=180°

Given, ∠CZY=30°

∠BYZ=180° −115° −30° =35°