A statue, 1.6 m tall, stands on the top of pedestal.

Solution:

In the figure, DC represents the statue and BC represents the pedestal.

Now, in right $\triangle \mathrm{ABC}$, we have

$\frac{\mathbf{A B}}{\mathbf{B C}}=\cot 45^{\circ}=1$

$\Rightarrow \frac{\mathbf{A B}}{\mathbf{h}}=1$

$\Rightarrow \mathrm{AB}=\mathrm{h}$ metres.

Now in right $\triangle \mathrm{ABD}$, we have

$\frac{\text { BD }}{\text { AB }}=\tan 60^{\circ}=\sqrt{\mathbf{3}}$

$\Rightarrow \mathrm{BD}=\sqrt{\mathbf{3}} \times \mathrm{AB}=\sqrt{\mathbf{3}} \times \mathrm{h}$

$\Rightarrow \mathrm{h}+1.6=\sqrt{\mathbf{3}} \mathrm{h}$

$\Rightarrow \mathrm{h}(\sqrt{\mathbf{3}}-1)=1.6$

$\Rightarrow h=\frac{1.6}{\sqrt{3}-1}=\frac{1.6}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$\Rightarrow h=\frac{1.6}{3-1} \times(\sqrt{3}+1)=\frac{1.6}{2} \times(\sqrt{3}+1)$

$=0.8(\sqrt{\mathbf{3}}+1) \mathrm{m}$

Thus, the height of the pedestal is $0.8(\sqrt{3}+1) \mathrm{m}$.