In an isosceles triangle ABC, 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 ∠ABC=75
∘
and ∠ACB=75
∘
.
The angle bisectors of ∠ABC and ∠ACB meet at O.

Therefore, in △BOC:

∠OBC=
2
1
​
∠ABC=
2
75
∘

​
=37.5
∘

∠OCB=
2
1
​
∠ACB=
2
75
∘

​
=37.5
∘

The sum of angles in △BOC is 180
∘
:

∠BOC+∠OBC+∠OCB=180
∘

∠BOC+37.5
∘
+37.5
∘
=180
∘

∠BOC+75
∘
=180
∘

∠BOC=180
∘
−75
∘
=105
∘