Hello, lalji!

We must understand "bearings" and make a good sketch . . .

A submarine is being followed by two ships, $\displaystyle A\text{ and }B$,

3.8 km apart, with $\displaystyle A$ due east of $\displaystyle B.$

If $\displaystyle A$ is on a bearing of 165° from the submarine

and $\displaystyle B$ is on the bearing of 205° from the submarine,

find the distance from the submarine to both ships.

Answers: 5.4 km and 5.7 km Code:

N
:
: 90°
S* - - - E
* * 75°
* 40° *
a * * b
* *
* 65° 75° *
B * * * * * * * A
3.8

We have: .$\displaystyle AB = 3.8$

Since $\displaystyle \angle NSA = 165^o,\text{ then: }\,\angle ESA = \angle A = 75^o$

Since major $\displaystyle \angle NSB = 205^o\!:\;\angle ASB = 40^o$

In $\displaystyle \Delta BSA\!:\;\angle B \:=\:180^o - 40^o - 75^o \:=\:65^o$

Let $\displaystyle a = SB,\;b = SA$

Law of Sines: .$\displaystyle \frac{a}{\sin75^o} \:=\:\frac{3.8}{\sin40^o} \quad\Rightarrow\quad a \:=\:5.710312527$

Law of Sines: .$\displaystyle \frac{b}{\sin65^o} \:=\:\frac{3.8}{sin40^o}\quad\Rightarrow\quad b \:=\:5.357865551$

Therefore: .$\displaystyle \boxed{\begin{array}{ccc}\text{A-to-S}&\approx & 5.4\text{ km} \\ \text{B-to-S}&\approx&5.7\text{ km} \end{array}}$