Question 868991
No such triangles exist (draw it out!). We can prove it using law of sines:


*[tex \large \frac{a}{\sin A} = \frac{b}{\sin B}]


Substituting a = 15, b = 30, A = 45 deg, we obtain *[tex \large \sin B = \sqrt{2}]. However sin B must be in [-1, 1]! Hence no triangle exists, and the problem you wrote has a mistake.