Question

A water skier is pulled behind a boat by a rope. The rope has a tension of 650 N and is at an angle of 27°. What is the y-component of the tension?

Answer 1

Answer: 295N

Explanation:

Answer 2

There is no figure, so I asume the angle of the rope is measured with respect to the horizontal direction.

If this is the case, then the magnitude of the tension is the hypothenuse of a right triangle, of which $T_x$ and $T_y$ (horizontal and vertical component of the tension) are the other sides, and $\alpha$ is the angle between T and $T_x$. 
Therefore, $T_y$ (the side of the triangle opposite to $\alpha$) is given by the magnitude of the tension multiplied by the sine of the angle:
$T_y = T \sin \alpha = (650 N)(\sin 27^{\circ} )=295 N$