Given: circle k(O), m AB =(6x−50)°, m∠BCA=(2x)° Find: m∠BAC
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Given: circle k(O), m AB =(6x−50)°, m∠BCA=(2x)° then m∠BAC = 40°
Step-by-step explanation:
∠AOB = 2∠BCA
=> (6x-50)° = 2(2x)°
=> (6x-50)° = (4x)°
=> (2x)° = 50°
∠BCA + ∠BAC + ∠ABC = 180° ( Sum of Angles of a trinagle)
∠ABC = 90°
(2x)° + ∠BAC + 90° = 180°
=> 50° + ∠BAC = 90°
=> ∠BAC = 40°
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