Question 165593
You have done ok so far!
In your triangle, you have {{{a = 1}}}, {{{c = 2}}}, and {{{b = sqrt(3)}}}.
To see how we get b, you have:
{{{b^2 = 3}}} Take the square root of both sides.
{{{b = sqrt(3)}}}
The sin of theta (I'll call it A because I can't do a theta symbol) is given as {{{1/2}}} because the sine of the angle is the side opposite the angle (a = 1) over the hypotenuse (c = 2), so you get {{{sin(A) = 1/2}}}
The cosine of this angle (A) is the side adjacent to the angle ({{{b = sqrt(3)}}} over the hypotenuse (c = 2), so...
{{{cosin(A) = sqrt(3)/2}}}