Hello, xwrathbringerx!

The elevation of a hill at a place $\displaystyle P$ due east of it is 48°

and at a place due south of $\displaystyle P$ the elevation is 30°.

If the distance from $\displaystyle P$ to $\displaystyle Q$ is 500 m, find the height of the hill.

Standing on the ground and looking north, we have this view: Code:

A *
| *
|42°*
h | *
| *
| *
| 48° *
B * - - - - - - * P

The hill is: .$\displaystyle h = AB.$

$\displaystyle \angle APB = 48^o \quad\Rightarrow\quad \angle BAP = 42^o$

. . Hence: .$\displaystyle BP = h\tan42^o$

Standing on the ground and looking southwest, we have this view: Code:

* A
* |
*60°|
* | h
* |
* |
* 30° |
Q * - - - - - - * B

The hill is: .$\displaystyle h = AB.$

$\displaystyle \angle AQB = 30^o \quad\Rightarrow\quad \angle QAB = 60^o$

. . Hence: .$\displaystyle BQ = h\tan60^o$

Looking down at the ground, we have this view: Code:

h·tan42°
B * - - - - - - * P
* |
* |
* | 500
h·tan60° * |
* |
* |
* Q

Using Pythagorus: .$\displaystyle \left(h\tan60^o\right)^2 \;=\;\left(h\tan42^o\right)^2 + 500^2 \quad\Rightarrow\quad h^2\tan^2\!60^o \;=\;h^2\tan^2\!42^o + 500^2 $

. . $\displaystyle h^2\tan^2\!60^o - h^2\tan^2\!42^o \:=\:500^2 \qquad\Rightarrow\qquad h^2\left(\tan^2\!60^o - \tan^2\!42^o\right) \:=\:500^2$

. . $\displaystyle h^2 \;=\;\frac{500^2}{\tan^2\!60^o - \tan^2\!42^o} \qquad\Rightarrow\qquad h \;=\;\frac{500}{\sqrt{\tan^2\!60^i - \tan^2\!42^o}}$ . . . . *There!*