Height of the tower

Labelling the width of the river as $d$ and the height of the tower as $h$, as shown below, we can then use trigonometry.

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From the triangle containing the 30$^\text{o}$ angle, $\tan{30}=\dfrac{h}{d}\Rightarrow d\tan{30}=h$.

From the triangle containing the 24$^\text{o}$ angle, $\tan{24}=\dfrac{h}{d+36}\Rightarrow (d+36)\tan{24}=h$.

These are simultaneous equations in $h$ and $d$ (since $\tan{30}$ and $\tan{24}$ are just numbers, which can be found using a calculator at any point).

Solving by substitution

The aim is to get $h$ on its own, so making $d$ the subject of the first equation gives $d=\dfrac{h}{\tan{30}}$.

Substituting this into the second equation gives $$\begin{align}&\left(\dfrac{h}{\tan{30}}+36\right)\tan{24}=h\\

\Rightarrow&\frac{h\tan{24}}{\tan{30}}+36\tan{24}=h\\

\Rightarrow&36\tan{24}=h-h\dfrac{\tan{24}}{\tan{30}}\\

\Rightarrow&36\tan{24}=h\left(1-\dfrac{\tan{24}}{\tan{30}}\right)\\

\Rightarrow&36\tan{24}=h\left(\dfrac{\tan{30}-\tan{24}}{\tan{30}}\right)\\

\Rightarrow&36\tan{24}\dfrac{\tan{30}}{\tan{30}-\tan{24}}=h\\

\Rightarrow &70.041=h\end{align}$$

So the tower is 70 metres tall.

Finding $d$ then $h$

If $d\tan{30}=h$ and $(d+36)\tan{24}=h,$ then $d\tan{30}=(d+36)\tan{24}.$ So $d\tan{30}-d\tan{24}=36\tan{24}$, so $d=\dfrac{36\tan{24}}{\tan{30}-\tan{24}}=121.31.$

And $d\tan{30}=h,$ so $h=\tan{30}\times 121.31=70.041$. So the tower is 70 metres tall.