Bearing problem.

EbenezerSon · 2013-07-21 01:15:00

An aeroplane flew from town A, travels 200km at a bearing of 135 digree to town B. Then travels 250km from town B at a bearing of 045 digree to town C. Find using the vector approach
(a) the distance of A from C.
(b) the bearing of A from C.

Please solve this problem for me.
Thanks in advance.

Last edited by EbenezerSon (2013-07-21 02:38:02)

Bob · 2013-07-21 04:11:14

stay on line I'm just making a diagram.

bob

I've got broadband connection problems so I hope I can do this

Bob · 2013-07-21 04:21:29

OK here's the diagram below.

Bearings are measured clockwise from north. The dotted lines are all showing north.

So AB has components East = 200sin(45) and North = -200cos(45)

get BC similarly.

Then add the easts and add the norths

Pythag will get the size of CA and arctan will get the angle. (convert it to a bearing). From my diagram it looks to be about 250 or thereabouts.

Bob

EbenezerSon · 2013-07-21 20:56:35

I had:

(a) 320km.
(b) 263 digrees.
I think I am right. I used vector approach to solve.

I tried using vector approach to solve the following, but got it wrong, please help.

A cyclist starts from a point A travel 8 km in the distance 030 digrees to B and then 10 km to C. Find the cyclist's distance and bearing from A.

Bob · 2013-07-21 21:25:35

hi EbenezerSon

Well done, those are my answers too. ( I rounded 263.65980825409 up to 264 )

For the next problem, you haven't said what the bearing is for stage two ( B --> C ) of the journey.

Bob

EbenezerSon · 2013-07-21 21:36:08

This how I solved it:

vector AB = (8km, 030 digrees).

Vector BC = (10km, 000).
I use 000 digrees, as the bearing from B to C, because it not provided.

Vector AB = (8cos60) = 8 * 0.5 = 4.
(8sin60) = 8 * 0.8660 = 6.928

Vector BC = (10 cos000) = 10 * 1 = 10.
(10sin000) = 10*0= 0.

To find distance, Vector AC = AB + BC.
AC = 4 + 10 = 14.
6.928 + 0 = 6.928.
By Pythag.
AC^2 = 14^2 + 6.928^2
=243.99. taking root
=15.620.

Bearing from North I had, 153.671.

Please, help me.

EbenezerSon · 2013-07-21 21:43:57

There several examples in the book, which distances and bearings was not provided, one example is what above.

Bob · 2013-07-21 21:49:22

You have to be told the second bearing so it looks like the book forgot to tell you this.

If you provide the book answer I'll work back and figure out the missing information.

Bob

EbenezerSon · 2013-07-21 21:54:10

There are several examples in the book, which there is no distance and bearing of locations provided,
But the book has the answer by using trigonometry to solve, but the book says it could be solved by vector approch as well. I will post another example of such problem.

Bob · 2013-07-21 23:05:03

Vector AB = (8cos60) = 8 * 0.5 = 4.
(8sin60) = 8 * 0.8660 = 6.928
Vector BC = (10 cos000) = 10 * 1 = 10.
(10sin000) = 10*0= 0.

60 is the angle with an easterly direction. 0 is the angle with a northerly direction. So you have muddled the components.

The easterly components are 4 and zero

The northerly components are 6.928 and 10

So you have added the wrong components together.

Your assumption that bearing C from B is zero appears to be correct.

Bob

EbenezerSon · 2013-07-22 08:29:24

Yes, I have had it right! I missed it up then. I would post more in which distances and bearings are not given.

Thanks Sir!

EbenezerSon · 2013-07-22 08:38:56

This is it:

A cyclist starts a journey from town A . He rides 10 km north, then 5 km east and finally ten 10 km on a bearing of 045 degrees

a) How far east is the cyclist's destination from A?

b) How far north is the cyclist's destination from town A?

c) Find the distance and the bearing of the cyclist's destination from A?. ( correct your answer to the nearest km and degree).

This is the problem.

Thanks. God bless.

Bob · 2013-07-22 09:41:18

OK. So, do you have an answer?

Bob

EbenezerSon · 2013-07-22 23:33:45

My final answers are different from what the book has. I used vector approach though. This is the respective answers from the book.

a) The cyclist's destination is 12 km east of town A.

b) The cyclist's destination is 17 km north of town A.

c) The cyclist's final destination is 21 km away on a bearing of 35 degrees from A.

But have you worked it out?

Last edited by EbenezerSon (2013-07-22 23:48:33)

Bob · 2013-07-23 00:19:56

I was waiting for your answers. I have now done the calculations and I get the same.

Bob

EbenezerSon · 2013-07-23 01:04:19

But please, did you use the vector approach? If so, please show me the workings, that I could learn.
Please.

EbenezerSon · 2013-07-23 01:10:16

Please produce them here.
Thanks.

Bob · 2013-07-23 06:14:30

I did use the vector approach much like you. I haven't kept my working (on Excel) but the shot below shows my working to get the missing bearing on the previous question.

Bob

EbenezerSon · 2013-07-23 06:51:15

I used zeros for the unprovided bearings. And got every thing wrong. I don't know any method well, apart from the vector approach. I don't seem to understand your workings, but see how I did mine:

From A to north; vector AN = (10km 000)
From north to east; vector NE = (5km, 000)
From east to the end let say P, therefore: vector EP = (10km, 045).

I used zeros for the unknown bearings, am I right?

EbenezerSon · 2013-07-23 07:55:24

It seems my inability to know the bearings always cost me.

Bob · 2013-07-23 19:50:31

hi EbenezerSon

Let's try to get this clear. Bearings are always measured clockwise from North. I have made a diagram (below) with point B on a bearing of x from A

East vector component NB = AB sin(x)
North vector component AN = AB cos(x)

If x is over 90, the sine will still be positive but the cosine will now come out negative.
If x is between 180 and 270, the sine will now be negative and so will the cosine.
If x is over 270, the sine will be negative and the cosine will once more be positive.

Calculators will automatically put the correct sign on the sine and cosine so you can always use these two formulas
Westerly components will automatically come out as negative Easterly and similarly South will be negative North.

So get all the components as Easts and Norths. Adding will always work because the negative signs will cause you to subtract those.

After this you can get the final distance with

and the bearing angle with

Unfortunately, the bearing may not come out correctly from the atan because (for example) atan(1) may be 45 or it may be 225. There is no way the calculator can 'know' which answer to use; so you'll have to look at the diagram to decide on the correct bearing. eg. If you get 45 and you can see the answer should be SouthWest you can correct your answer by adding 180.

In your example

From A to north; vector AN = (10km 000)
From north to east; vector NE = (5km, 000)
From east to the end let say P, therefore: vector EP = (10km, 045).

first stage East = 0 North = 10
second stage East = 5 North = 0
third stage East = 10sin(45) = 7.07 North = 10cos(45) = 7.07

totalE = 12.07 total North = 17.07

distance = root(12.07^2 + 17.07^2) = 20.907....

angle = atan(12.07/17.07 = 35.26....

Bob

EbenezerSon · 2013-07-23 23:55:51

So should I always use east and north as my guide?.

I am trying hard to understand what you mean.

Last edited by EbenezerSon (2013-07-24 00:00:51)

Bob · 2013-07-24 00:11:03

Yes, use East and North always, then you'll start to develop a method.

Bob

Math Is Fun Forum

#1 2013-07-21 01:15:00

Bearing problem.

#2 2013-07-21 04:11:14

Re: Bearing problem.

#3 2013-07-21 04:21:29

Re: Bearing problem.

#4 2013-07-21 20:56:35

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#5 2013-07-21 21:25:35

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#6 2013-07-21 21:36:08

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#7 2013-07-21 21:43:57

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#8 2013-07-21 21:49:22

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#9 2013-07-21 21:54:10

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#10 2013-07-21 23:05:03

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#11 2013-07-22 08:29:24

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#12 2013-07-22 08:38:56

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#13 2013-07-22 09:41:18

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#14 2013-07-22 23:33:45

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#15 2013-07-23 00:19:56

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#16 2013-07-23 01:04:19

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#17 2013-07-23 01:10:16

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#18 2013-07-23 06:14:30

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#19 2013-07-23 06:51:15

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#20 2013-07-23 07:55:24

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#21 2013-07-23 19:50:31

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#22 2013-07-23 23:55:51

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#23 2013-07-24 00:11:03

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