Question

find the angle (theta) using two heights in a triangle
(picture posted below)

Answer 1

\[\tan \theta = \frac{h}{x} = \frac{h'}{x+L}\]
since i assume "x" is unknown, solve for "x" and substitute
\[x = \frac{h}{\tan \theta}\]
then
\[\large \tan \theta = \frac{h'}{\frac{h}{\tan \theta}+L}\]
solve for tan
\[\tan \theta = \frac{h' \tan \theta}{h+L \tan \theta}\]
\[1 = \frac{h'}{h+L \tan \theta}\]
\[\tan \theta = \frac{h'-h}{L}\]
finally
\[\theta = \tan^{-1} (\frac{h'-h}{L})\]

Answer 2

oh wow haha , just noticed an easier way without all the algebra

using the smaller proportional triangle you also get
\[\tan \theta = \frac{h'-h}{L}\]