(i) C = 40°
(ii) ∠BOC = 120°
Using angle sum property of a triangle in ∆ABC,
We have
→ ∠A + ∠B + ∠C = 180°
→ 60° + 80° + ∠C = 180°
→ ∠C = 180° - 60° - 80°
→ ∠C = 40°
Now,
Since OB and OC are the bisectors of ∠B and ∠C respectively, thus
Using angle sum property of a triangle in ∆BOC,
→ ∠BOC + ∠OBC + ∠OCB = 180°
→ ∠BOC + ∠B/2 + ∠C/2 = 180°
→ ∠BOC + 80°/2 + 40°/2 = 180°
→ ∠BOC + 40° + 20° = 180°
→ ∠BOC = 180° - 40° - 20°
→ ∠BOC = 120°
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Answer :
(i) C = 40°
(ii) ∠BOC = 120°
Concept to be used :
Solution :
Using angle sum property of a triangle in ∆ABC,
We have
→ ∠A + ∠B + ∠C = 180°
→ 60° + 80° + ∠C = 180°
→ ∠C = 180° - 60° - 80°
→ ∠C = 40°
Now,
Since OB and OC are the bisectors of ∠B and ∠C respectively, thus
Now,
Using angle sum property of a triangle in ∆BOC,
We have
→ ∠BOC + ∠OBC + ∠OCB = 180°
→ ∠BOC + ∠B/2 + ∠C/2 = 180°
→ ∠BOC + 80°/2 + 40°/2 = 180°
→ ∠BOC + 40° + 20° = 180°
→ ∠BOC = 180° - 40° - 20°
→ ∠BOC = 120°
Hence,
∠C = 40°, ∠BOC = 120°