Bob

]]>I am trying hard to understand what you mean.

]]>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

]]>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

]]>Thanks.]]>

Please.]]>

Bob

]]>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

]]>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.

]]>Thanks Sir!

]]>Vector AB = (8cos60) = 8 * 0.5 = 4.

(8sin60) = 8 * 0.8660 = 6.928Vector 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

]]>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.]]>