We are given BC = 16, AC = 5, and the measure of angle B = 42°.
Applying the law of sines:
[tex]\frac{16}{sin\text{ A}}=\frac{5}{sin\text{ B}}[/tex]
Where A is the angle in vertex A.
Solving for sin A:
[tex]sin\text{ A}=\frac{16\text{ }sin\text{ 42\degree}}{5}=\frac{16*0.669}{5}=2.14[/tex]
We have the sine of angle A with a value that is greater than 1. The sine can only have values between -1 and 1, so this triangle has no solution.
