Question

In ∆ABC, Angle A=60°, angle B=80° and the bisector of angleB and angleC meet at O. Find angle C and angle BOC

Answer 1

It is easy follow my steps
make a triangle roughly , then name the angle at left be angle c and angle at right be angle b also angle at top be angle a
Now draw two line from b/w angle b and c so they meet at a common point o
ans:
Given: angle a=60°
angle b=80°
To find: angle c and angle BOC
Solution: angle a+b+c =180°[angle sum property of triangle ]
60+80+c=180
c=180-140
c=40
therefore angle c =40
angle boc + angle bco + angle cbo = 180°(Angle sum property.....)
angle boc + 20 + 40 =180
angle boc = 180-60
angle boc = 120°
I have taken half values in above equation because ob or oc bisect both the angles into 2 equal halves

even the word bisect means dividing an angle into 2 equal halves

Answer 2

Answer:

Measure of angle C is 40° and angle BOC is 120°.

Step-by-step explanation:

Given a triangle ABC.

The measure of angles, $\angle A=60^o$ and $\angle B=80^o$

Applying the sum angle property of a triangle, that is the sum of all the three interior angles in a triangle is 180°.

Hence, in $\triangle ABC$ we have $\angle A + \angle B + \angle C = 180^o$

$\implies 60^o + 80^o + \angle C= 180^o$

$\implies 140+ \angle C= 180$

$\implies \angle C= 180-140=40$

Therefore, the measure of angle C is 40°.

Given the bisectors of angle B and angle C meet at O. This make a $\triangle OBC$.

Bisector of angle B gives,

$\angle OBC =\dfrac{\angle B}{2} =\dfrac{80}{2}=40^o$

And the bisector of angle C gives,

$\angle OCB =\dfrac{\angle C}{2} =\dfrac{40}{2}=20^o$

In $\triangle OBC$, $\angle OCB + \angle OBC +\angle BOC = 180^o$

$\implies 20^o + 40^o+ \angle BOC = 180^o$

$\implies 60+ \angle BOC = 180$

$\implies \angle BOC = 180- 60 = 120$

Therefore, the measure of angle BOC is 120°.

Therefore, angle C = 40° and angle BOC = 120°.