# Bearings question with trig.

• May 10th 2009, 06:39 AM
mitchoboy
Bearings question with trig.
Hey Im really not sure how to do bearings at all.
For homework i have this question:

A ship leaves at port A and travels for 30km on bearing of 120degrees
It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
Calculate distance AB and bearing A from B

thanks
• May 10th 2009, 07:42 AM
aidan
Quote:

Originally Posted by mitchoboy
...
A ship leaves at port A and travels for 30km on bearing of 120degrees
It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
Calculate distance AB and bearing A from B
...

Typically bearings are given from a reference [North or South] and deflecting East or West.

From the information given, assume the reference for zero bearing is due North or the y-axis; and assume that port A is at the origin (0,0).

$\displaystyle X_0 = 0$
$\displaystyle Y_0 = 0$

$\displaystyle X_1 = X_0 + \sin \left(120\right) \times\ 30 = 0.866 \times 30 = 25.981$
$\displaystyle Y_1 = Y_0 + \cos \left(120\right) \times\ 30 = 0.500 \times 30 = 15.000$

$\displaystyle X_2 = X_1 + \sin\left( 80\right) \times\ 50 = X_1 + 0.985 \times 50 = X_1 + 49.240 = 75.221$
$\displaystyle Y_2 = Y_1 + \cos \left(80\right) \times\ 50 = Y_1 + 0.174 \times 50 = Y_1 + 8.682 = 23.682$

Since $\displaystyle X_0 = 0$ & $\displaystyle Y_0 = 0$ The distance AB is $\displaystyle \sqrt{X_2^2 + Y_2^2}$

The bearing is the arctangent of the difference between final coordinates and the initial coordinates:

The tangent of the bearing AB is : $\displaystyle \frac{X_2 - X_0} {Y_2 - Y_0}$

As a check:

$\displaystyle X_2 = \sin \left ({bearing AB}\right) \times \left ({distance AB}\right)$

$\displaystyle Y_2 = \cos \left ({bearing AB}\right) \times \left({distance AB}\right)$
• May 10th 2009, 07:42 AM
galactus
Just for kicks, let's do it this way. Let's use a coordinate system so you can see what is going on. Like in surveying.

Let's say the coordinate of A is (0,0).

Then, to the turning point, it is 30 km at an azimuth of 120 degrees.

(Technically, this is an azimuth, not a bearing. But, it doesn't really matter).

$\displaystyle x=30sin(120)=25.98$

$\displaystyle y=30cos(120)=-15$

Those are the coordinates of the turning point, (25.98,-15)

Next, the boat turns 80 degrees from north and goes 50 km to go to B.

B's coordinates are $\displaystyle x=25.98+50sin(80)=75.22$

$\displaystyle y=-15+50cos(80)=-6.32$

The coordinates of B are (75.22, -6.32).

Now, to find the distance back to A where it started, just use ol' Pythagoras.

$\displaystyle \sqrt{(75.22)^{2}+(-6.32)^{2}}=75.485$

To find the bearing back to A, one way of many:

$\displaystyle 270+cos^{-1}(\frac{75.22}{75.485})=274.8 \;\ deg$

Here is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice.
• May 10th 2009, 08:26 AM
Soroban
Hello, mitchoboy!

Bearings are measured clockwise from North.
And a good diagram is essential.

Quote:

A ship leaves at port $\displaystyle A$ and travels for 30 km on bearing of 120°.
It then changes course and travels for 50 km on bearing of 080° arriving at port $\displaystyle B.$
Calculate distance $\displaystyle AB$ and bearing of $\displaystyle {\color{blue}A}$ from $\displaystyle {\color{blue}B}.$ .
Is this correct?
Code:

      N       |       |     A o                    R       | *    *            |       |60°*          *    |       |    *      Q      o B       |    30 *    |    *       S        *60°|80°* 50                   * | *                     o                     P

The ship starts at A and sails 30 km to point $\displaystyle P$:
. . $\displaystyle AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

Then it turns and sails 50 km to point $\displaystyle B$:
. . $\displaystyle PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$

In $\displaystyle \Delta APB$, use the Law of Cosines:

. . $\displaystyle AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$

. . $\displaystyle AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$

Therefore: .$\displaystyle \boxed{AB \;\approx\;75.5\text{ km}}$

In $\displaystyle \Delta APB$, use the Law of Cosines.

. . $\displaystyle \cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$

Hence: .$\displaystyle \angle A \;\approx\;25.2^o$

Then: .$\displaystyle \angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$

Therefore, the bearing of $\displaystyle A$ from $\displaystyle B$ is: .$\displaystyle 360^o - 85.2^o \:=\:\boxed{274.8^o}$

• May 11th 2009, 09:02 AM
mitchoboy
Quote:

Originally Posted by Soroban
Hello, mitchoboy!

Bearings are measured clockwise from North.
And a good diagram is essential.

Code:

      N       |       |     A o                    R       | *    *            |       |60°*          *    |       |    *      Q      o B       |    30 *    |    *       S        *60°|80°* 50                   * | *                     o                     P
The ship starts at A and sails 30 km to point $\displaystyle P$:
. . $\displaystyle AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

Then it turns and sails 50 km to point $\displaystyle B$:
. . $\displaystyle PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$

In $\displaystyle \Delta APB$, use the Law of Cosines:

. . $\displaystyle AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$

. . $\displaystyle AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$

Therefore: .$\displaystyle \boxed{AB \;\approx\;75.5\text{ km}}$

In $\displaystyle \Delta APB$, use the Law of Cosines.

. . $\displaystyle \cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$

Hence: .$\displaystyle \angle A \;\approx\;25.2^o$

Then: .$\displaystyle \angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$

Therefore, the bearing of $\displaystyle A$ from $\displaystyle B$ is: .$\displaystyle 360^o - 85.2^o \:=\:\boxed{274.8^o}$

thankyou for your answer. but how did you get apq as 60 degrees?