# Thread: Trigonometry Question with bearings

1. ## Trigonometry Question with bearings

I need help on text book question: (I am confused, I think that enough information is not given) please help me if you know how solve the problem.

Question:
A submarine is being followed by two ships, A and B, 3.8 km apart, with A due east of B. If A is on a bearing of 165 degree from the submarine and B is on the bearing of 205 degree from the submarine, find the distance from the submarine to both ships.

Answer given at the back of text book is: 5.4km and 5.7km

2. Measure "bearing from the submarine" from the direction the submarine is travelling and with reference to a viewer on the submarine..

Bearing 0º from the submarine would be directly in front.

Bearing 180º from the submarine would be directly in behind.

Bearing 90º from the submarine would be directly to the starboard.

Bearing 270º from the submarine would be directly to the port.

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

4. ## Thanks for the response

You have explained it very well. Thank you very much.