Question

in figure, Bp bisects angle abc and ab=ac.find x

Answer 1

in ABC ab=ac
so, angle ABC = angle ACB (angle opposite to equal sides are equal)
let angle ABC to be x
so x+x+angle bac = 180
2x+ 60= 180
x= 60
we have
angle abp = angle cbp
so, angle cbp= 30
we have ap//bc
so, angle APB =angle cbp
x=30

Answer 2

AB = AC, and BP bisects ∠ABC

AP ∥ BC is drawn

Now ∠PBC = ∠PBA

(∵ PB is the bisector of ∠ABC)

∴ AP||BC

∴ ∠APB = ∠PBC (Alternate angles)

⇒ x = ∠PBC …(i)

In ∆ ABC, ∠A = 60°

and ∠B = ∠C (∵ AB = AC)

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 60° + ∠B + ∠C = 180°

⇒ 60° + ∠B + ∠B = 180°

⇒ 2∠B = 180° - 60° = 120°

∠B = 120°/2 = 60°

= 1/2∠B = 60°/2 = 30° ⇒ ∠PBC = 30°

∴ From (i)

x= 30°