Hello, prashantvrm!

They used "nice" angles in these problems.

We can solve them without a calculator.

. . I'll walk through the first one.

1) As observed from the top of a 100 m lighthouse,

the angles of depression of two ships approaching it are 30° and 45°.

If one ship is directly behind the other, find the distance between the two ships. Code:

A * - - - - - - - - - - - C
| * * 30°
| * *
| * *
100 | * *
| * *
| 45° * 30° *
* - - - - - - * - - - - - - *
B y P x Q

The lighthouse is $\displaystyle AB = 100.$

The ships are at $\displaystyle P$ and $\displaystyle Q.$

$\displaystyle \angle CAQ = 30^o$, hence: $\displaystyle \angle AQB = 30^o$

$\displaystyle \angle CAP = 45^o$, hence: $\displaystyle \angle APB = 45^o$

Let $\displaystyle x = PQ,\;y = BP$

In $\displaystyle \Delta ABP\!:\;\tan45^o \:=\:\frac{y}{100} \quad\Rightarrow\quad\frac{y}{100} \:=\:1 \quad\Rightarrow\quad y \:=\:100 $ .[1]

In $\displaystyle \Delta ABQ\!:\;\;\tan30^o \:=\:\frac{100}{x+y} \quad\Rightarrow \quad \frac{1}{\sqrt{3}} \:=\:\frac{100}{x+y} \quad\Rightarrow\quad x+y \:=\:100\sqrt{3}$ .[2]

Substitute [1] into [2]: .$\displaystyle x + 100 \:=\:100\sqrt{3}$

. . Therefore: .$\displaystyle x \;=\;100\sqrt{3} - 100 \;=\;100(\sqrt{3}-1) $