At the foot of a mountain, the elevation of it summit is 45°;

Solution:

Suppose, AB is a mountain of height t + x.

In $\triangle D F C$,

$\sin 30^{\circ}=\frac{x}{1000}$

$\Rightarrow x=1000 \times\left(\frac{1}{2}\right)=500 \mathrm{~m}$

and,

$\tan 30^{\circ}=\frac{x}{y}$

$\Rightarrow y=\frac{x}{\tan 30^{\circ}}=500 \sqrt{3}$

In $\Delta A B C$,

$\tan 45^{\circ}=\frac{t+x}{y+z}$

$\Rightarrow t+x=y+z \quad \ldots(1)$

In $\Delta A D E$

$\tan 60^{\circ}=\frac{t}{z}$

$\Rightarrow t=\sqrt{3} z \quad \ldots(2)$

From (1) and (2), we have

$\sqrt{3} z+x=y+z$

$\Rightarrow z(\sqrt{3}-1)=500(\sqrt{3}-1)$

$\Rightarrow z=500 \mathrm{~m}$

$\therefore t=\sqrt{3} z=500 \sqrt{3}$

Hence, height of the mountain $=t+x=500 \sqrt{3}+500=500(\sqrt{3}+1) \mathrm{m}$.