Hello, iiharthero!

A slight variation of Aidan's excellent approach . . .

A man walking due north observes a church spire with bearing 342°.

After walking 1500m, he observes it with bearing 337°.

How far is the church from the road?

I'll use Aidan's diagram . . .

Code:

h
C * - - - - - - - - - * D
* * |
* * | x
* * 23° |
* * |
* * B
* |
* 18° | 1500
* |
* |
* A

The church is at $\displaystyle C.$ .The man walks along the road $\displaystyle AD.$

He makes his first observation at $\displaystyle A\!:\;\angle A = 18^o$

He walks 1500 m to $\displaystyle B \!:\;AB = 1500$

. . And $\displaystyle \angle CBD = 23^o.$

The distance from the church to the road is: $\displaystyle h = CD.$

. . Let $\displaystyle x = BD.$

In right triangle $\displaystyle CDB\!:\;\;\tan23^o \:=\:\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan23^o}$ .[1]

In right triangle $\displaystyle CDA\!:\;\;\tan18^o \:=\:\frac{h}{x+1500} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan18^o} - 1500$ .[2]

Equate [2] and [1]: .$\displaystyle \frac{h}{\tan18^o}-1500 \:=\:\frac{h}{\tan23^o}$

Multiply by $\displaystyle \tan18^o\tan23^o\!:\quad h\tan23^o - 1500\tan18^o\tan23^o \:=\:h\tan18^o$

. . and we have: .$\displaystyle h\tan23^o - h\tan18^o \:=\:1500\tan18^o\tan23^o$

Factor: . $\displaystyle h(\tan23^o -\tan18^o) \:=\:1500\tan18^o\tan23^o$

Therefore: .$\displaystyle h \;=\;\frac{1500\tan18^o\tan23^o}{\tan23^o-\tan18^o} \;\approx\;2078\text{ m}$