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bob bundy
2013-07-24 22:11:03

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

Bob

EbenezerSon
2013-07-24 21:55:51

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

I am trying hard to understand what you mean.

bob bundy
2013-07-24 17:50:31

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.

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
2013-07-24 05:55:24

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

EbenezerSon
2013-07-24 04:51:15

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?

bob bundy
2013-07-24 04:14:30

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 23:10:16

Thanks.

EbenezerSon
2013-07-23 23:04:19

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

bob bundy
2013-07-23 22:19:56

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

Bob

EbenezerSon
2013-07-23 21:33:45

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?

bob bundy
2013-07-23 07:41:18

OK.  So, do you have an answer?

Bob

EbenezerSon
2013-07-23 06:38:56

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.

EbenezerSon
2013-07-23 06:29:24

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!

bob bundy
2013-07-22 21:05:03

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 19:54:10

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.