If O be the centre of incircle of a triangle ABC and ∠BOC = 110°, then the…
2019
If O be the centre of incircle of a triangle ABC and ∠BOC = 110°, then the value of ∠BAC will be
- A.
110°
- B.
20°
- C.
40°
- D.
55°
Show answer & explanation
Correct answer: C
For a triangle ABC with incenter O, O lies on the internal bisectors of all three angles, so ∠OBC = B/2 and ∠OCB = C/2. Since the angles of triangle OBC sum to 180°, this gives the standard incenter relation ∠BOC = 90° + A/2, where A = ∠BAC.
Write the incenter relation: ∠BOC = 90° + (∠BAC)/2.
Substitute the given value ∠BOC = 110°: 110° = 90° + (∠BAC)/2.
Isolate the fraction: (∠BAC)/2 = 110° − 90° = 20°.
Solve for the full angle: ∠BAC = 2 × 20° = 40°.
Cross-check by substituting back: with ∠BAC = 40°, the incenter relation gives ∠BOC = 90° + 40°/2 = 90° + 20° = 110°, which matches the given value, confirming ∠BAC = 40°.