How does author derives $\cos \theta = \frac{dx}{ds} \qquad \sin\theta = \frac{dy }{ds}, $?

Question

In the book of Dynamics by Horace Lamb, at page 103, is it given

that for a motion on a smooth curve, the equation of motion is given by $$mv \frac{dv }{ ds} = -mg \sin \theta \qquad \frac{ mv^2 }{r} = -mg \cos\theta + R, $$ where $\theta$ is the angle between the surface normal and vertical direction, and $R$ is the "pressure" exerted by the curve.

[...]

Then, we have $$\cos \theta = \frac{dx}{ds} \qquad \sin\theta = \frac{dy }{ds}, $$ where $s$ is length of the path taken over the surface, and $x,y$ are usual cartesian coordinates as $y$ is taken as upward.

However, I cannot understand how does the author derives the latter equations between $\theta$ and the derivatives of $x,y$ wrt $s$.

Answer 1

Hint Write it as $dx=ds\cos\theta$ and $dy=ds\sin\theta$. It is just a decomposition of $ds$ into components along $x$ and $y$ axes

Answer 2

Now apply the rule of $\sin \theta$ and $\cos \theta$