Geometry Practice Test 7

Construction - from point O draw a perpendicular bisector of BC.
⇒ ∠ODB = 90°
In ΔODB,
∠BOD+∠ODB+∠DBO = 180°
55° + 90° +∠DBO = 180°
∠DBO = 35°
⇒ ∠B = 70°
Similarly, we find ∠C = 70°

In ΔABC,
∠A +∠B+∠C = 180°
∠BAC = 180° − (70° + 70°)
∠BAC = 40°