The weight \(W = mg\) is neither along the +x or the +y directions. But, it has both x- and y-components. To calculate this, note that triangle ABC is a right-angled triangle, as angle ACB is \(90 ^{\circ}\). Angle ABC is thus \(90 - \theta\)

Also, as the x-axis is a straight line, we have the \(180^{\circ}\). From all this, we see that the vector \(\vec{W}\) makes an angle of \(180 + 90 - \theta = 270 - \theta\) with the +x-axis.

We know that \(W_x\) is given by the \(\cos\) component of the angle the vector makes with the +x-axis. Thus, \(W_x = W \cos \left(270 - \theta \right) = - W \sin \theta\). Similarly, \(W_y = - W \cos \theta\). (Note the minus signs! It is because +x and +y are upwards, and weight is acting downwards.)