Question

Q 26). In the given figure BO, CO are the bisectors of external angles CBE and BCD respectively of triangle ABC, angle BAC is 60 degree then angle BOC is equal to​

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Answer 1

Step-by-step explanation:

As BO and CO are the angle bisectors of external angles of△ABC, Then

∠1=∠2

∠4=∠3

We know, ∠A+∠ABC+∠ACB=180 ∘

…eqn(1)

And ∠ABC=180−2∠1

∠ACB=180−2∠4

Putting it in the eqn (1), we get

∠A+180−2∠1+180−2∠4=180

⇒∠1+∠4=90+ 21 ∠A…eqn(2)

Also we know from the figure, ∠BOC+∠1+∠4=180∘

∠BOC=180−∠1−∠4

From eqn (2)

∠BOC=180−90− 21 ∠A

⇒∠BOC=90 ∘ − 21 ∠A

Answer 2

Step-by-step explanation:

As BO and CO are the angle bisectors of external angles of△ABC, Then

∠1=∠2∠4=∠3

We know, ∠A+∠ABC+∠ACB=180∘…eqn(1)

And ∠ABC=180−2∠1∠ACB=180−2∠4

Putting it in the eqn (1), we get

∠A+180−2∠1+180−2∠4=180⇒∠1+∠4=90+21∠A…eqn(2)

Also we know from the figure, ∠BOC+∠1+∠4=180∘

∠BOC=180−∠1−∠4

From eqn (2)

∠BOC=180−90−21∠A⇒∠BOC=90∘−21∠A