Length of the shadow of a tower at 4:00 pm compared to the length of the shadow at 10:00 am Consider a location on the Earth where the Sun is overhead at noon. Compared to its shadow at 10:00 am, the shadow of a tower at 4:00 pm would be
(1) twice longer
(2) three times longer
(3) four times longer
(4) eight times longer
Let $h$ be the height of the tower. Let $x,y$ be the lengths of the shadow at 10:00 am and 4:00 pm, respectively. If $\theta$ is the angle of elevation of the Sun from the ground at 10:00 am, then the corresponding angle of elevation at 4:00 pm should be $\theta/2$, as the Sun is overhead at noon (12:00 noon).
Now,
\begin{align}
\tan\theta=\frac{h}{x}, \quad \tan\frac{\theta}{2}=\frac{h}{y}.
\end{align}
Also,
\begin{align}
\tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}=\frac{2\cdot\frac{h}{y}}{1-\frac{h^2}{y^2}}.
\end{align}
So, we have
\begin{align}
\frac{h}{x}=\frac{2\cdot\frac{h}{y}}{1-\frac{h^2}{y^2}}\\
\implies\frac{1}{x}=\frac{2}{y-\frac{h^2}{y}}.
\end{align}
What to do after that?

You would compare the two after you calculate it.
Here is another solution:

$\frac{y}{x} = tan(60) = \sqrt 3$.
Therefore, $y = \sqrt 3 \times x$.

$\frac{z}{x}=tan(30)=\frac{1}{\sqrt3}$
Therefore, $z = \frac{x}{\sqrt 3}$.
$\frac{y}{z}=\frac{x \times \sqrt 3}{\frac{x}{\sqrt 3}} = 3$.
Therefore, the shadow will be 3 times longer.
