Consider the following figure:

In $∆\mathrm{AEC}$,

$\tan \left({30}^{°}\right) = \frac{\mathrm{AE}}{\mathrm{EC}}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{\mathrm{AE}}{120}$

$\therefore \mathrm{AE} = 69.28\mathrm{m}$

In $∆\mathrm{BEC}$,

$\tan \left({24}^{°}\right) =\frac{\mathrm{EB}}{\mathrm{EC}}$

$\Rightarrow 0.445 = \frac{\mathrm{EB}}{120}$

$\therefore \mathrm{EB} = 53.427\mathrm{m}$

Thus, the height of first tower, AB = AE + EB

                                                  = 69.282 + 53.427

                                                  = 122.709m = 122 m

And height of second tower, CD = EB = 53.427m = 53.4m

Let consider A be the position of the aeroplane and let BC be the river. Let D be the point in BC just below the aeroplane.

For ΔADC, we get

$\tan \left(45°\right) = \frac{\mathrm{h}}{\mathrm{y}}$

$\Rightarrow 1 = \frac{250}{\mathrm{y}}$

$\Rightarrow \mathrm{y} = 250\mathrm{m}$

For ΔADB, we get

$\tan \left(60°\right) = \frac{\mathrm{AD}}{\mathrm{DB}}$

$\Rightarrow \sqrt{3} = \frac{\mathrm{h}}{\mathrm{x}}$

$\Rightarrow \sqrt{3} = \frac{250}{\mathrm{x}}$

$\Rightarrow \mathrm{x} = \frac{250}{\sqrt{3}}\mathrm{m}$

Therefore, the width of the river is DB + DC = 250 + $\frac{250}{\sqrt{3}}$ = 394m.

Let  PQ  be the light house.

$$\begin{gathered} \Rightarrow \mathrm{PQ}\hspace{0.17em}= 60 \\ \mathrm{In} ∆\mathrm{PQA}, \\ \tan 60° = \frac{\mathrm{PQ}}{\mathrm{AQ}} \\ \Rightarrow \sqrt{3} = \frac{60}{\mathrm{AQ}} \\ \Rightarrow \mathrm{AQ} = \frac{60}{\sqrt{3}} \\ \Rightarrow \mathrm{AQ} = \frac{20 \mathrm{X} 3}{\sqrt{3}} \\ \Rightarrow \mathrm{AQ} = \frac{20 \mathrm{X} \sqrt{3} \mathrm{X} \sqrt{3}}{\sqrt{3}} \\ \Rightarrow \mathrm{AQ} = 20 \sqrt{3} \mathrm{m}. \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{PQB}, \\ \tan 45° = \frac{\mathrm{PQ}}{\mathrm{QB} } \\ \Rightarrow 1 = \frac{60}{\mathrm{QB}} \\ \Rightarrow \mathrm{QB} = 60 \mathrm{m} \\ \mathrm{Now}, \\ \mathrm{AB} = \mathrm{AQ} + \mathrm{QB} \\ = 20 \sqrt{3} + 60 \end{gathered}$$

      = 20  x  1.732  +  60 

      = 94.64

      = 95 m.

A   is the aeroplane, D  and  C  are the ships sailing towords  A.

Ships are sailing towords the aeroplane in the same direction.

in the figure, height  Ab = 1500 m.

To find:  Distance between the ships, that is  CD.

Solution: 

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{right}-\mathrm{angled} ∆\mathrm{ABC}, \\ \tan 45° = \frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1 = \frac{1500}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC} = 1500 \mathrm{m}. \\ \mathrm{In} \mathrm{the} \mathrm{right}-\mathrm{angled} ∆\mathrm{ABD}, \\ \tan 30° = \frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500}{\mathrm{BD}} \\ \Rightarrow \mathrm{BD} = 1500 \sqrt{3} \\ \Rightarrow \mathrm{BD} = 1500 ( 1.732 ) \\ = 2598 \mathrm{m} \\ \therefore \mathrm{Distance} \mathrm{between} \mathrm{the} \mathrm{ships} = \mathrm{CD} = \mathrm{BD} - \mathrm{BC} \\ = 2598 - 1500 \\ = 1098 \mathrm{m}. \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{PQR} \\ \tan 60° = \frac{\mathrm{RQ}}{\mathrm{PQ}} \\ \sqrt{3} = \frac{50}{\mathrm{PQ}} \\ \mathrm{PQ}= \frac{50}{\sqrt{3}} \\ \mathrm{In} ∆\mathrm{PQT} \\ \tan 30° = \frac{\mathrm{PT}}{\mathrm{PQ}} \\ \frac{1}{\sqrt{3}} = \frac{\mathrm{PT}}{{\displaystyle \frac{50}{\sqrt{3}}}} \\ \mathrm{PT} = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} \\ \mathrm{PT} = \frac{50}{3}. \end{gathered}$$