How is $\arctan2(\sqrt{h}, \sqrt{1-h}) = \arcsin(\sqrt{h})$?

Question

According to this link here: http://www.movable-type.co.uk/scripts/latlong.html

Hiversine distance is: $2 \times r \times \arctan2(\sqrt{h}, \sqrt{1-h})$

Where $r$ is radius of the Earth.

Where as on this link: https://en.wikipedia.org/wiki/Haversine_formula#:~:text=The%20haversine%20formula%20determines%20the,and%20angles%20of%20spherical%20triangles.

Hiversine distance is: $2 \times r \times \arcsin(\sqrt{h} )$

so

$\arctan2(\sqrt{h}, \sqrt{1-h}) = \arcsin(\sqrt{h})$

How is this true?

Since:

$\arcsin(x) = 2 \arctan(\dfrac{x}{1 + \sqrt{1-x^2}})$

We would have

$\arcsin(\sqrt{h}) = 2\arctan(\dfrac{\sqrt{h}}{1 + \sqrt{1-h}})$

How do I go from here to $\arctan2(\sqrt{h}, \sqrt{1-h})$?

Many thanks in advance!

Answer 1

It can be seen in a right triangle with sides $\sqrt{1-h}$, $\sqrt{h}$ and (by the Pythagorean theorem) $1$.