# Find $\phi_2$ given $d, \phi_1$ and $\lambda_1 = \lambda_2$

Suppose that you have have two points $$x = (\phi_1, \lambda_1)$$ and $$y = (\phi_2, \lambda_2)$$, where $$\phi$$ and $$\lambda$$ represent latitude and longitude. Let $$\Delta\phi = \phi_2 - \phi_1$$ and $$\Delta\lambda = \lambda_2 - \lambda_1$$, then the distance between the two points in meters, $$d$$, is given by the following system

\begin{align} &a = \sin²(\Deltaφ/2) + \cos φ_1 ⋅ \cos φ_2 ⋅ \sin²(\Delta\lambda/2)\\ &c = 2 ⋅ atan2( \sqrt{a}, \sqrt{(1−a)} )\\ &d = R ⋅ c \end{align}

or alternatively by

$$d = cos^{-1}( \sin \phi_1 ⋅ \sin \phi_2 + \cos \phi_1 ⋅ \cos \phi_2 ⋅ \cos \Delta\lambda ) ⋅ R\tag{1}$$

(Source for these equations can be found here)

Question: Suppose that $$\lambda_1 = \lambda_2$$, and suppose that I know $$\phi_1$$. How do I determine $$\phi_2$$ such that the distance between $$(\phi_1, \lambda_1)$$ and $$(\phi_2, \lambda_2)$$ is $$d$$?

What I've tried: If $$\lambda_1 = \lambda_2$$ then equation $$(1)$$ above can be reduced to $$\cos\big( \frac{d}{R}\big) = \sin\phi_1\sin\phi_2\\ \Leftrightarrow \phi_2 = sin^{-1}\bigg(\dfrac{\cos(d/R)}{\sin(\phi_1)}\bigg)$$ This doesn't work and I would like to know why!

I believe you have mistakenly reduced the expression $$\cos{\phi_1}\cos{\phi_2}\cos{\Delta \lambda}$$ to $$0$$ instead of $$1$$, as $$\cos{0}=1$$. Then we have $$\cos{\frac{d}{R}} = \cos{(\phi_2-\phi_1)}$$ $$\phi_2 = \frac{d}{R}+\phi_1+2k\pi, k \in \mathbb{Z}$$