Bearings question with trig.

mitchoboy · #1 $Old$ May 10th, 2009, 08:39 AM

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

aidan · #2 $Old$ May 10th, 2009, 09:42 AM

Quote:

Originally Posted by mitchoboy $View Post$

...
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).

$X_0 = 0$
$Y_0 = 0$

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

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

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

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

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

As a check:

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

$Y_2 = \cos \left ({bearing AB}\right) \times \left({distance AB}\right)$

galactus · #3 $Old$ May 10th, 2009, 09:42 AM

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).

$x=30sin(120)=25.98$

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

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 $x=25.98+50sin(80)=75.22$

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

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

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

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

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

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

Here is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice.

Soroban · #4 $Old$ May 10th, 2009, 10:26 AM

Hello, mitchoboy!

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

Quote:

A ship leaves at port $A$ 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 $B.$
Calculate distance $AB$ and bearing of ${\color{blue}A}$ from ${\color{blue}B}.$ . 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 $P$ :
. . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

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

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

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

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

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

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

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

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

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

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

mitchoboy · #5 $Old$ May 11th, 2009, 11:02 AM

Quote:

Originally Posted by Soroban $View Post$

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 $P$ :
. . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

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

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

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

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

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

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

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

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

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

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

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