Question 1067578
The picture:
{{{n=sin(theta/2+alpha/2)/sin(theta/2)}}} {{{drawing(300,250,-3,3,-2.5,2.5,
triangle(-2,-1.732,2,-1.732,0,1.732),
red(arc(0,1.732,1,1,60,120)),
line(-3,-1.578,-1,0),line(4,-2.367,1,0),
line(-1,0,1,0),arrow(-1,0,0.3,0),
arrow(-3,-1.578,-2,-0.789),locate(-0.1,1.6,red(alpha)),
arrow(1,0,2,-0.789),green(arrow(0,0.789,-4,-2.367)),
green(arrow(0,0.789,2,2.367)),green(arrow(0,0.789,4,-2.367)),
green(arc(0,0.789,2,2,-41,41)),locate(1.1,0.9,green(theta))
)}}}
{{{alpha=60^o}}}= prism angle crossed by the light
{{{alpha/2=60^o/2=30^o}}}
{{{theta}}}= angle of deviation of the light as it goes through the prism
{{{sin(alpha/2)=sin(30^o)=1/2=0.5}}}
{{{cos(alpha/2)=cos(30^o)=sqrt(3)/2=0.5sqrt(3)}}}

 
To calculate {{{sin(theta/2+alpha/2)}}} we need the trigonometric identity formula for a sum of angles:
{{{sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B)}}} .
 
{{{sin(theta/2+alpha/2)=sin(theta/2+30^o)=sin(theta/2)*cos(30^o)+cos(theta/2)*sin(30^o)=0.5sqrt(3)*sin(theta/2)+0.5*cos(theta/2)}}}
So substituting the expression above,
and the value {{{1.5}}} for {{{n}}}
in the formula for refractive index, we get
{{{1.5=(0.5sqrt(3)*sin(theta/2)+0.5*cos(theta/2))/sin(theta/2)}}} .
Multiplying both sides of the equal sign times {{{2sin(theta/2)}}},
we get the equivalent equation
{{{3sin(theta/2)=sqrt(3)*sin(theta/2)+cos(theta/2)}}}
Now, dividing both sides of the equation by {{{cos(theta/2)}}} ,
we get the equivalent equation
{{{3sin(theta/2)/cos(theta/2)=sqrt(3)*sin(theta/2)/cos(theta/2)+cos(theta/2)/cos(theta/2)}}} ,
which simplifies to
{{{3tan(theta/2)=sqrt(3)*tan(theta/2)+1}}} ,
so
{{{3tan(theta/2)-sqrt(3)*tan(theta/2)=1}}} --> {{{(3-sqrt(3))*tan(theta/2)=1}}} --> {{{tan(theta/2)=1/(3-sqrt(3))}}}
From
{{{tan(theta/2)=1/(3-sqrt(3))}}} , or the more elegantly written equivalent
{{{tan(theta/2)=(3+sqrt(3))/6}}} ,
we can calculate an approximate value for {{{tan(theta/2)}}} ,
for which your calculator would give you an approximate value for {{{theta/2}}} ,
which can be used to calculate an approximate value for {{{theta}}} .
The approximate values would be:
{{{tan(theta/2)=0.788675}}} , {{{theta/2=38.26^o}}} ,
and {{{highlight(theta=76.52^o)}}} .
 
An exact value would have to be expressed as an inverse tangent function
{{{theta=2*tan^(-1)((3+sqrt(3))/6)}}} ,
or if you use the double angle trigonometric identity to calculate that {{{tan(theta)=(30+14sqrt(3))/13}}} ,
{{{theta=tan^(-1)((30+14sqrt(3))/13)}}} .
Hopefully no instructor would make you go through all that trouble.