Question

Given: circle k(O), m AB =(6x−50)°, m∠BCA=(2x)° Find: m∠BAC

Answer 1

Answer:

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°