This triangle is not a right triangle. How do we solve this then? You will use the law of sine with is shown below:
![(sin A)/(a) =(sin B)/(b) = (sinC)/(c)](https://img.qammunity.org/2020/formulas/mathematics/high-school/8n6tvl3tq4pvav4rlvvrnokdr1a5ty2107.png)
What we know is shown in the image attached below:
Plug what you know into the law of sine
![(sin75)/(51.2) =(sinB)/(33.7)](https://img.qammunity.org/2020/formulas/mathematics/high-school/yip7trybezba50bevdv4fd0qknrqkc2j0x.png)
To solve for sinB cross multiply
sin75*33.7 = sinB * 51.2
32.55 = sinB*51.2
Divide 51.2 to both sides to isolate sinB
32.55 / 51.2 = sinB / 51.2
0.63577 = sinB
To find B you must use arcsin:
![sin^(-1) 0.63577](https://img.qammunity.org/2020/formulas/mathematics/high-school/uqh8albat70nymke8c3yfij62gbhp0dq6w.png)
39.477
^^^This is your rough estimate but you can simply keep it to 39 degrees
This means that your answer is correct!
Hope this helped!