The angles of depression of the top and bottom of a tower as seen from the top of a 60

Solution:

Let AD be the tower and BC be the cliff.

We have,

$\mathrm{BC}=60 \sqrt{3} \mathrm{~m}, \angle \mathrm{CDE}=45^{\circ}$ and $\angle \mathrm{BAC}=60^{\circ}$

Let $\mathrm{AD}=h$

$\Rightarrow \mathrm{BE}=\mathrm{AD}=h$

$\Rightarrow \mathrm{CE}=\mathrm{BC}-\mathrm{BE}=60 \sqrt{3}-h$

In $\Delta \mathrm{CDE}$,

$\tan 45^{\circ}=\frac{\mathrm{CE}}{\mathrm{DE}}$

$\Rightarrow 1=\frac{60 \sqrt{3}-h}{\mathrm{DE}}$

$\Rightarrow \mathrm{DE}=60 \sqrt{3}-h$

$\Rightarrow \mathrm{AB}=\mathrm{DE}=60 \sqrt{3}-h \quad \ldots$ (i)

Now, in $\triangle \mathrm{ABC}$,

$\tan 60^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}$

$\Rightarrow \sqrt{3}=\frac{60 \sqrt{3}}{60 \sqrt{3}-h} \quad[\mathrm{Using}(\mathrm{i})]$

$\Rightarrow 180-h \sqrt{3}=60 \sqrt{3}$

$\Rightarrow h \sqrt{3}=180-60 \sqrt{3}$

$\Rightarrow h=\frac{180-60 \sqrt{3}}{\sqrt{3}}$

$\Rightarrow h=\frac{180-60 \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow h=\frac{180 \sqrt{3}-180}{3}$

$\Rightarrow h=\frac{180(\sqrt{3}-1)}{3}$

$\therefore h=60(\sqrt{3}-1)$

$=60(1.732-1)$

$=60(0.732)$

Also, $h=43.92 \mathrm{~m}$

So, the height of the tower is 43.92 m.