Explanation:

In Figure , AB is the tower and $\mathrm{BC}$ is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and $\mathrm{DB}$ is the length of the shadow, when the angle of elevation is 30°.
Now, let $\mathrm{AB}$ be $h\mathrm{m}$ and $\mathrm{BC}$ be $x\mathrm{m}$. According to the question, $\mathrm{DB}$ is $40\mathrm{m}$ longer than $\mathrm{BC}$.
So, $\mathrm{DB}=(40+x)\mathrm{m}$
Now, we have two right triangles $\mathrm{ABC}$ and $\mathrm{ABD}$.
In ΔABC, tan60°= $ \displaystyle \dfrac{{AB}}{{BC}}$
or ,  √3 = $\displaystyle \dfrac{h}{x}$
In ΔABD, tan30° = $ \displaystyle \dfrac{{AB}}{{BD}}$
​i.e.,  $\displaystyle \dfrac{1}{{\sqrt{3}}}=\dfrac{h}{{x+40}}$
Putting this value in (2), we get (x√3​) √3​ = x + 40,
i.e., 3x = x + 40
i.e., x = 20
So, h = 20√3 ​[From(1)]
Therefore, the height of the tower is 20√3​ m.