# Calculate the height of a building

This question I really need help with, I simply do not know where to start! Anyone can help, all I can offer is supreme thanks. Please include method. I don't want simple answers which don't help me learn

Let the length of legs of the small right triangle be $h,x$, then use trigonometry to get two relations including these unknowns
$$\frac{h}{20+x}=\tan 23 \\ \frac{h}{x}=\tan 39$$

Let $h$ be the height of building and $d$ be the distance from $Y$ to the bottom of building. Therefore, we have the following relations: $$\tan 39^\circ=\frac hd\quad\Rightarrow\quad d=\frac h{\tan 39^\circ}\tag1$$ and $$\tan 23^\circ=\frac h{d+20}\quad\Rightarrow\quad d+20=\frac h{\tan 23^\circ}.\tag2$$ Substitute $d$ in $(1)$ to $(2)$ yield \begin{align} \frac h{\tan 39^\circ}+20&=\frac h{\tan 23^\circ}\\ 20&=\frac h{\tan 23^\circ}-\frac h{\tan 39^\circ}\\ 20&=h\left(\frac 1{\tan 23^\circ}-\frac 1{\tan 39^\circ}\right)\\ 20&=h\left(\frac {\tan 39^\circ-{\tan 23^\circ}}{\tan 23^\circ{\tan 39^\circ}}\right)\\ \Large\color{blue}h&=\Large\color{blue}{\frac{20\tan 23^\circ{\tan 39^\circ}}{\tan 39^\circ-{\tan 23^\circ}}}. \end{align}

Let the point falling down from Z be T and the distance from Y to T be m

Apply tanQ in triangle XZT and triangle YZT which will give

tan 23 = x/(20+m) & tan 39 = x/m

\begin{align} &\tan 39^\circ=h/w \implies w=\frac{h}{\tan 39^\circ} \; \tag{*}\\ &\tan 23^\circ=\frac{h}{(20+w)} \implies 20\tan23^\circ+w\tan23^\circ=h \; \tag{#}\\ &\text{Hence from * and \#,}\;\; 20\tan23^\circ+\tan23^\circ\times\left({{h}\over{\tan39^\circ}}\right)=h\\ &\rightarrow h=(20\tan23^\circ)\cdot \left(1-\frac{\tan23^\circ}{\tan39^\circ}\right)=4.039..\approx 4.04 \;\rm{m} \end{align}
1. identify what you want to solve for (in this case the height $h$)