The figure shows a light ray incident on a polished metal cylinder where h = R/6. At what angle θ will the ray be reflected?
1 Answer

This is as much about geometry as physics.
Draw a construction line from the centre of the circle through the point where the ray hits the cylinder. That line is the normal against which the angle of incidence and angle of reflection are measured. { The reflection surface is the tangent to the cylinder at the point where the ray strikes }
We know that angle of incidence = angle of reflection = θ / 2
Drop a vertical construction line from the point where the ray hits the cylinder to the horizontal line through the centre of the cylinder. Its length = h.
You have a right angled triangle with a hypotenuse of length R
The angle at the centre between the horizontal line and the 'normal' construction line is equal to the angle of incidence of the ray {corresponding angles on parallel lines.}
h / R = sin(θ/2)
h = R/6
R / 6R = sin(θ/2)
1/6 = sin(θ/2)
θ/2 = arcsin(1/6) = 9.59°
θ = 19.19°