cos(x) = sqrt(8)/3
not sure how you're supposed to find this, but one way is to use the trigonometric identity of sin^2(x) + cos^2(x) = 1.
this becomes (1/3)^2 + cos^2(x) = 1.
subtract (1/3)^2 from both sides of this equation to get:
cos^2(x) = 1 - (1/3)^2
simplify to get:
cos^2(x) = 1 - 1/9 which then becomes:
cos^2(x) = 9/9 - 1/9 which then becomes:
cos^2(x) = 8/9).
take the square root of both sides of this equation to get:
cos(x) = sqrt(8)/3.
you can also find it by using the right triangle as your guide.
let your right triangle be ABC with the right angle at C.
a diagram of your triangle is shown below:
in this diagram, you can see that side a is opposite angle A and side b is opposite angle B and side c is opposite angle C.
sine (A) = opposite / hypotenuse = a/c = 1/3.
this means that a is equal to 1 and c is equal 3.
since this is a right triangle, a^2 + b^2 = c^2 which translates to:
1^2 + b^2 = 3^2 which simplifies to:
1 + b^2 = 9
subtract 1 from both sides of this triangle to get:
b^2 = 8 which makes:
b = sqrt(8).
cosine (A) = adjacent / hypotenuse = b/c = sqrt(8) / 3.
you get the same answer either way.