Question

The bisects of exterior angles at b and c of triangle abc meet at o. If angle a = x , then angle boc = ?

Answer 1

Answer:

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Answer 2

∠ BOC  =  90° + x / 2

Given: The bisectors of exterior angles at b and c of Δ ABC meet at O.

The angle a = x.

To Find: The ∠ BOC

Solution:

In Δ BOC,

∠ OBC + ∠OCB + ∠ BOC = 180°                               ........ (1)

In Δ ABC,

  ∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 2 × ∠OCB + 2 × ∠ BOC = 180° [From (1), since they are bisectors]

⇒ ∠ A / 2 + ∠OCB + ∠ BOC = 90°

⇒ ∠OCB + ∠ BOC = 90° - ∠ A / 2                   ....... (2)

Putting (2) in (1), we get;

    90° - ∠ A / 2 + ∠ BOC = 180°      

⇒  ∠ BOC = 90° + ∠ A / 2

                 = 90° + x / 2                          [ since  ∠ A = x ]

Hence,  ∠ BOC  =  90° + x / 2

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