In the diagram, the size of $\angle A C B$ is
(A) $57^{\circ}$
(B) $37^{\circ}$
(C) $47^{\circ}$
(D) $60^{\circ}$
(E) $17^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-076.jpg?height=312&width=577&top_left_y=1939&top_left_x=1292)
Since $\angle A B C+\angle A B D=180^{\circ}$ (in other words, $\angle A B C$ and $\angle A B D$ are supplementary) and $\angle A B D=130^{\circ}$, then $\angle A B C=50^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-334.jpg?height=304&width=574&top_left_y=1751&top_left_x=838)

Since the sum of the angles in triangle $A B C$ is $180^{\circ}$ and we know two angles $93^{\circ}$ and $50^{\circ}$ which add to $143^{\circ}$, then $\angle A C B=180^{\circ}-143^{\circ}=37^{\circ}$.

ANswer: (B)