In an isosceles triangle, the angle bisectors of ∠B and ∠C intersect at point O. Find the angle ∠BOC (in degrees), when ∠ABC=∠ACB=75



∘
.

In △ABC, we are given that ∠ABC=75
∘
and ∠ACB=75
∘
.

Since OB is the angle bisector of ∠ABC:

∠OBC=
2
∠ABC
​
=
2
75
∘

​
=37.5
∘

Since OC is the angle bisector of ∠ACB:

∠OCB=
2
∠ACB
​
=
2
75
∘

​
=37.5
∘

Now, consider △BOC. The sum of angles in a triangle is 180
∘
:

∠BOC+∠OBC+∠OCB=180
∘

∠BOC+37.5
∘
+37.5
∘
=180
∘

∠BOC+75
∘
=180
∘

∠BOC=180
∘
−75
∘
=105
∘