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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 551

Hi,

Please, I wanted to know if all bearing problems could be solved by using vector approach method. Please confirm

Many thanks!

I know only one thing - that is that I know nothing

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**Olinguito****Member**- Registered: 2014-08-12
- Posts: 649

Yes.

*Bassaricyon neblina*

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 551

Thank you. I will put a bearing question here which I think cannot be solved by vector appraoch. As far as I remember, Bob bundy made it clear to me that not all bearing questions could be solved using vector approach, I wanted to know this truth from different perspective. Thanks once more!

Pece kololo!

*Last edited by EbenezerSon (2015-05-18 05:00:00)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

As far as I remember, Bob Bundy made it clear to me that not all bearing questions could be solved using vector approach

Did I say that ? I've searched old posts and found only one on this topic. I did say then that your calculator would not sort out the correct angle without you referring to a diagram, but the approach still used vectors. I look forward to the question.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 551

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

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

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

(C) Find the distance and bearing of the cyclist's destination from town A.

(Correct your answers to nearest kilometers and degree)

Is hard for me using vector approach

I started by saying

(10km, 000°) (5km, 000°) (10km, 045°) I used Zero degrees for the unknown degrees.

(10cos000° and 10sin000°) =(10*cos90 and 10*sin 90) = (0 And 10) I did the same for the rest.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

hi EbenezerSon

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

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

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

(C) Find the distance and bearing of the cyclist's destination from town A.

(Correct your answers to nearest kilometers and degree)

Total Easterly component = 0 + 5 + 10sin(45)

Total Northerly component = 10 + 0 + 10cos(45)

This should work out OK. Post answers please.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 551

Hi; bob thank you - but are you using vector approach for this question?

Thank you

*Last edited by EbenezerSon (2015-05-19 01:29:23)*

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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 551

So, if I am getting it right does it mean, the 45 is the bearing for all the kilometers travelled by the cyclist? Please confirm.

I know only one thing - that is that I know nothing

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,371

hi EbenezerSon

My apologies for the late reply. I hadn't realised you had replied to my post.

but are you using vector approach for this question?

Yes because I have calculated the two vector components of the journey.

45 is the bearing for all the kilometers travelled by the cyclist?

No. The journey is in three parts. One part is North; one part is East; only part three is on a bearing of 45.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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