ABC is a triangle in which ∠A = 72°, the internal bisectors of angl

1.	ABC is a triangle in which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Answer» Given, ABC is a triangle ∠A = 72° and internal bisectors of B and C meet O. In ΔABC ∠A + ∠B + ∠C = 180° 72° + ∠B + ∠C = 180° ∠B + ∠C = 180° – 72° ∠B + ∠C = 108° Divide both sides by 2, we get \(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = \(\frac{108}{2}\) \(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = 54° ∠OBC + ∠OCB = 54° (i) Now, in ΔBOC ∠OBC + ∠OCB + ∠BOC = 180° 54° + ∠BOC = 180°[Using (i)] ∠BOC = 180° – 54° = 126°

ABC is a triangle in which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.

Answer»

Given,

ABC is a triangle

∠A = 72° and internal bisectors of B and C meet O.

In ΔABC

∠A + ∠B + ∠C = 180°

72° + ∠B + ∠C = 180°

∠B + ∠C = 180° – 72°

∠B + ∠C = 108°

Divide both sides by 2, we get

\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = \(\frac{108}{2}\)

\(\frac{∠B}{2}\) + \(\frac{∠C}{2}\) = 54°

∠OBC + ∠OCB = 54° (i)

Now, in ΔBOC

∠OBC + ∠OCB + ∠BOC = 180°

54° + ∠BOC = 180°[Using (i)]

∠BOC = 180° – 54°

= 126°