The correct answer is:

20âˆš3

20âˆš3

**Explanation:**

In Figure , AB is the tower and $BC$ is the length of the shadow when the Sun’s altitude is $6_{Â°}$, i.e., the angle of elevation of the top of the tower from the tip of the shadow is $6_{Â°}$ and $DB$ is the length of the shadow, when the angle of elevation is $3_{Â°}$.

Now, let $AB$ be $hÂm$ and $BC$ be $xÂm$. According to the question, $DB$ is $40Âm$ longer than $BC$.

So, $DB=(40+)m$

Now, we have two right triangles $ABC$ and $ABD$.

In $Î”ABC,tan6_{ABD,tanABBDâ€‹i.e.,Â $\displaystyle \dfrac{1}{{\sqrt{3}}}=\dfrac{h}{{x+40}}$(xâˆš3â€‹) âˆš3â€‹=+Â= xÂ+x=h=âˆš3â€‹[From()]âˆš3â€‹Âm}$