Question

In Fig. 10, triangle ABC is an isosceles triangle where AB = AC. BO and CO are the bisectors of ZABC and ZACB respectively. Find the value of the angles of triangle ABC.


Please answer fastly.. Plz​

Answer 1

Step-by-step explanation:

In ΔABC

AB=AC

∠ACB=∠ABC

2

1

∠ACB=

2

1

∠ABC

∠OCB=∠OBC

in ΔBOC

∠OBC+∠OCB+∠BDC=180

∘

2∠OBC+∠BOC=180

∘

∠ABC+∠BOC=180

∘

∠ABC+∠OBA=180

∘

∠DBA=∠BOC

Answer 2

Step-by-step explanation:

Given :-

∆ABC is an isosceles triangle where AB = AC. BO and CO are the bisectors of <ABC and <ACB respectively.

To find :-

Find the value of the angles of triangle ABC?

Solution :-

Given that

In∆ABC is an isosceles triangle ,AB = AC.

=> <ABC = <ACB

Since The angles opposite to the equal sides are equal.

Let <ABC = <ACB = X°

Now,

BO and CO are the bisectors of <ABC and <ACB

=> <ABC= <ABO+<OBC

and

<ABO= <OBC = <ABC /2 = X°/2

and

<ACB = <ACO + <OCB

and

<ACO = <OCB=<ACB /2 = X°/2

Now,

In ∆ OBC ,

<OBC + <BOC+ <BCO = 180°

Since The sum of all angles in a triangle is 180°

=> (X°/2)+100°+(X°/2) = 180°

=> (2X°/2)+100° = 180°

=> X° = 180°-100°

=> X° = 80°

=> <ABC = <ACB = 80°

Now

Now,

In ∆ABC,

<ABC + <BCA+ <CAB = 180°

Since The sum of all angles in a triangle is 180°

=> 80°+<BAC+80°= 180°

=> <BAC+160° = 180°

=> <BAC = 180°-160°

=> <BAC = 20°

Therefore,<ABC = <ACB = 80° and <BAC = 20°

Answer:-

The angles of the triangle are 20° ,80° and 80°

Used formulae:-

→ The sum of all angles in a triangle is 180°

→The angles opposite to the equal sides are equal.