Question 537190
You may not need to find a value of x to find the sine, cosine and tangent, but you do need more information than what you wrote. 
If a right triangle goes with those values, like this
{{{drawing (200,100, -3,23,-1.5,11.5,
  triangle( 0, 0, 17.3, 0, 17.3, 10 ), 
  locate( 8.2, 7, c ), locate( 18, 5, a ), locate( 9, 0, b ), 
  locate( -1, 0, A ), locate( 18, 0, C ), locate( 18, 11, B ))}}} you have enough information.
If they just tell you that ABC is a triangle with a right angle at C, it is the same thing, and you have enough information.
If you do not know that it is a right triangle, then there is no way to solve the problem. 
If you know it is a right triangle, but you do not know which angle is the right angle, then there could be more than one answer set, because c could be the hypotenuse, or it could be one of the legs of the right triangle.
I will assume it's a right triangle and side c=2x is the hypotenuse (as in my drawing).
The right angle, opposite the hypotenuse would be called C (with the same letter, but using capitals).
The angle opposed to side a is normally called A.
The other angle and its opposite side would be B and b respectively.
The trigonometric ratios are define as
sinA = opposite side/hypotenuse, so sinA = a/c
cosA = adjacent side/hypotenuse, so cosA = b/c, and
tanA = opposite side/adjacent side, so tanA = a/b
You know that sinA = a/c = x/2x = 1/2
If you'd already been taught that, you would know that means A =30 degrees or {{{pi/6}}},
and you would know the values for cosine and tangent of that angle.
Otherwise, you would have to calculate the length of side b using Pythagoras theorem to continue.
{{{a^2+b^2=c^2}}}--> {{{x^2+b^2=(2x)^2}}}--> {{{b^2=3x^2}}}-->{{{b=sqrt(3)x}}}
Then cosA = {{{b/c = sqrt(3)x/2x=sqrt(3)/2}}} and
tanA ={{{a/b=x/sqrt(3)x=1/sqrt(3)=sqrt(3)/3}}}