\(A = 60^\circ, a = 2, b = 10\)
Using the Law of Sines we get \(\dfrac{\sin B}{10}=\dfrac{\sin 60^\circ}{2}=\dfrac{\sqrt{3}}{4}\), so \(\sin B=\dfrac{5\sqrt{3}}{2}\approx4.33 > 1\).
Since the sine of an angle cannot exceed \(1\), there is no solution to this triangle.