In a Δ ABC, If ∠A = 60°, ∠B =80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC=
A. 60°
B. 120°
C. 150°
D. 30°
Answers
Given: In a Δ ABC, If ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O.
To Find : ∠BOC
Proof :
In ∆ABC,
Since Sum of the angles of a triangle is 180° :
∠A + ∠B + ∠C = 180°
60° + ∠B + ∠C = 180°
∠B + ∠C = 180° - 60°
∠B + ∠C = 120°
½ ∠B + ½ ∠C = ½ × 120°
½ ∠B + ½ ∠C = 60°
½ (∠B + ∠C ) = 60°………….. (1)
In ∆BOC,
Since Sum of the angles of a triangle is 180° :
∠BOC + ∠OBC + ∠OCB = 180°
∠BOC + ½ ∠B + ½ ∠C = 180°
∠BOC + ½ (∠B + ∠C) = 180°
∠BOC + 60° = 180°
[From eq (1)]
∠BOC + 60° = 180° - 60°
∠BOC = 120°
Hence , ∠BOC is 120°.
Among the given options option (B) 120° is correct.
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Similar questions :
In Δ ABC, ∠A=50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
A. 25°
B. 50°
C. 100°
D. 75°
https://brainly.in/question/15906908
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = 1/2 ∠A, then ∠A is equal to
A. 80°
B. 75°
C. 60°
D. 90°
https://brainly.in/question/15906902
Answer:
Step-by-step explanation:
Since Sum of the angles of a triangle is 180° :
∠A + ∠B + ∠C = 180°
60° + ∠B + ∠C = 180°
∠B + ∠C = 180° - 60°
∠B + ∠C = 120°
½ ∠B + ½ ∠C = ½ × 120°
½ ∠B + ½ ∠C = 60°
½ (∠B + ∠C ) = 60°………….. (1)
In ∆BOC,
Since Sum of the angles of a triangle is 180° :
∠BOC + ∠OBC + ∠OCB = 180°
∠BOC + ½ ∠B + ½ ∠C = 180°
∠BOC + ½ (∠B + ∠C) = 180°
∠BOC + 60° = 180°
[From eq (1)]
∠BOC + 60° = 180° - 60°
∠BOC = 120°