Geometry: path length in atmosphere ("round" Earth)

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I'm having trouble obtaining this physics formula. Since it's mostly about geometry, I hope it isn't out of place here.

I'll paste the text from the book:

Considering the curvature of the Earth (R is the Earth radius) and a nonvertical direction (zenith angle θ), the relation between h and path length L in the atmosphere is:

$$h = L \cos θ + \frac12\frac{L^2}{R} \sin^2 θ$$

h is the atmosphere's height.

I understand the first term (which due to the inclination) but I can't find a way to get the second term (which is introduced by considering the "roundness" of the Earth)

Any help would be greatly appreciated

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On the triangle formed by the point of entry of the light ray into the atmosphere (we assume the atmosphere is a sphere of finite radius), the point of observation and the Earth's center we can apply the law of cosines to find that

$$h=\sqrt{R^2+L^2+2LR\cos\theta}-R$$

We can Taylor expand this function in powers of $L/R$ assuming that $L\ll R$ (which means the Earth is very big compared to it's atmosphere so it's approximately flat). Keeping terms up to quadratic order we find

$$h=L\cos\theta+\frac{L^2}{2R}\sin^2\theta+\mathcal{O}\left(\frac{L^3}{R^2}\right)$$