The bisectors of exterior angles at B and C of Δ ABC meet at O, if �

1.	The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =A. 90°+ \(\frac{x°}{2}\)B. 90°- \(\frac{x°}{2}\)C. 180°+ \(\frac{x°}{2}\)D. 180°- \(\frac{x°}{2}\)
Answer» ∠OBC = 180° - ∠B - \(\frac{1}{2}\)(180° - ∠B) ∠OBC = 90° - \(\frac{1}{2}\)∠B And, ∠OCB = 180° - ∠C - \(\frac{1}{2}\)(180° - ∠C) ∠OCB = 90° - \(\frac{1}{2}\)∠C In ΔBOC ∠BOC + ∠OCB + ∠OBC = 180° ∠BOC + 90° - \(\frac{1}{2}\)∠C + 90° - \(\frac{1}{2}\)∠B = 180° ∠BOC = \(\frac{1}{2}\)(∠B + ∠C) ∠BOC = \(\frac{1}{2}\)(180° - ∠A) [From ] ∠BOC = 90° - \(\frac{1}{2}\)∠A ∠BOC = 90° - \(\frac{x}{2}\)

The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =A. 90°+ \(\frac{x°}{2}\)B. 90°- \(\frac{x°}{2}\)C. 180°+ \(\frac{x°}{2}\)D. 180°- \(\frac{x°}{2}\)

Answer»

∠OBC = 180° - ∠B - \(\frac{1}{2}\)(180° - ∠B)

∠OBC = 90° - \(\frac{1}{2}\)∠B

And,

∠OCB = 180° - ∠C - \(\frac{1}{2}\)(180° - ∠C)

∠OCB = 90° - \(\frac{1}{2}\)∠C

In ΔBOC

∠BOC + ∠OCB + ∠OBC = 180°

∠BOC + 90° - \(\frac{1}{2}\)∠C + 90° - \(\frac{1}{2}\)∠B = 180°

∠BOC = \(\frac{1}{2}\)(∠B + ∠C)

∠BOC = \(\frac{1}{2}\)(180° - ∠A) [From ]

∠BOC = 90° - \(\frac{1}{2}\)∠A

∠BOC = 90° - \(\frac{x}{2}\)