Step-by-step Solution:

The equation of the ellipse is: \[ \frac{x^2}{16} + \frac{y^2}{1} = 1 \] The slope of the tangent is: \[ \tan 60^\circ = \sqrt{3}. \] Let the equation of the tangent be: \[ y = mx + c, \] where \( m = \sqrt{3} \). For the tangent to the ellipse, the condition is: \[ c^2 = a^2m^2 + b^2, \] where \( a^2 = 16 \) and \( b^2 = 1 \). Substituting these values: \[ c^2 = 16 \cdot (\sqrt{3})^2 + 1 = 16 \cdot 3 + 1 = 48 + 1 = 49. \] Thus: \[ c = \pm \sqrt{49} = \pm 7. \] Therefore, the equations of the tangents are: \[ y = \sqrt{3}x + 7 \quad \text{and} \quad y = \sqrt{3}x - 7. \]