Respuesta :

Answer:

[tex] m\angle BCA= 33\degree[/tex]

Step-by-step explanation:

[tex] In\: \odot A[/tex] BC is tangent to the circle at point B.

Therefore, by tangent-radius theorem:

[tex] \therefore AB\perp BC[/tex]

[tex] \therefore \angle ABC = 90\degree [/tex]

[tex] In\: \triangle ABC, [/tex]

[tex] m\angle ABC +m\angle BAC +m\angle BCA= 180\degree [/tex]

[tex] 90\degree+57\degree +m\angle BCA= 180\degree [/tex]

[tex] 147\degree +m\angle BCA= 180\degree [/tex]

[tex] m\angle BCA= 180\degree - 147\degree[/tex]

[tex] m\angle BCA= 33\degree[/tex]

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