Find the number of distinct triangles with measurements of A=31 a = 10, and b = 40

i think doesn t exsist:

$\frac{10}{\sin{31}}=\frac{40}{sinB} \\ 10sin(B)=40\sin(31) \\ sin(B)=2.06015229964\\ B=?$

but probably i am wrong so wait for boss

You are correct, Solveit

Using the Law of Cosines.....we can find the possible angle opposite the longest side

cos-1  [  (40^2  - 31^2  - 10^2) / (-2*31*10) ]  =

cos-1 [ -0.8693548387096774] = about 150.38°

But.....this angle, added to the given one of 31°,  would  give us a triangle wose angle sum > 180......which is impossible.....