(a) The law of reflection says that the angle of incidence (measured with respect to the perpendicular to the interface between the two mediums) is equal to the angle of reflection (again, measured with respect to the interface between the two mediums):
[tex]\theta _i = \theta _r[/tex]
In our problem, [tex]47.5 ^{\circ}[/tex] is the angle of the incident ray with respect to the surface, so the angle of incidence (with respect to the perpendicular) is
[tex]\theta_i = 90^{\circ} - 47.5 ^{\circ} = 42.5^{\circ}[/tex]
And so, the angle of reflection is also [tex]\theta_r = 42.5^{\circ}[/tex], and since this is measured with respect to the perpendicular to the surface, the angle between the reflected ray and the surface is
[tex]\theta= 90^{\circ}-42.5^{\circ}=47.5^{\circ}[/tex]
(b) The angle of refraction, [tex]\theta_t[/tex], is given by Snell's law:
[tex]n_i \sin \theta_i = n_t \sin \theta_t[/tex]
where [tex]n_i = 1.00 [/tex] is the refractive index of air and [tex]n_t = 1.66 [/tex] is the refractive index of glass.
Substituting, we find:
[tex]\theta_T = \arcsin ( \frac{n_i}{n_r} \sin \theta_i )=\arcsin ( \frac{1.00}{1.66} \sin (42.5^{\circ}) )=24.0^{\circ}[/tex]
However, this is the angle of refraction, measured with respect to the perpendicular to the surface; so, the angle of the refracted ray with respect to the surface is
[tex]\theta=90^{\circ}-24.0^{\circ}=66^{\circ}[/tex]
