Question

A Triangle ABC has, angles A= 60, B=70. The incenter of this triangle is at I. Find the angle BIC.

Answer 1

Answer:

ANSWER

In △ABC,

∠ABC+∠ACB+∠BAC=180

∠ABC+70+50=180

∠ABC=60

∘

∠OCB=

2

1

(180−∠ACB)

∠OCB=

2

1

(180−50)

∠OCB=65

∘

∠OBC=

2

1

(180−∠ABC)

∠OBC=

2

1

(180−60)

∠OBC=60

∘

In △OBC,

∠OCB+∠OBC+∠BOC=180

65+60+∠BOC=180

∠BOC=180−125

∴∠BOC=55

∘

Answer 2

Geometry

As we know the angle sum property of a triangle states that the sum of all the angles of a triangle is $180\textdegree$.

$\angle A= 60\textdegree\\\\\angle B= 70\textdegree$

I is the incenter (incenter is a point where the three angle bisectors of a triangle meets.)

$\angle A+ \angle B+\angle C=180\textdegree\\\\\implies 60\textdegree+70\textdegree+\angle C=180\textdegree\\\\\implies \angle C= 50\textdegree$

So, considering ΔBIC, and applying angle sum property

$\angle B+\angle I+\angle C=180\textdegree\\\\\implies 70\textdegree+\angle I+50\textdegree=180\textdegree\\\\\implies \angle I=60\textdegree$

Hence $\angle BIC=60\textdegree$.