Need help on Trigonometry; Bearings problem (1 Viewer)

[Aeyis]

Hi, I just have a question from homework and need some help on a question, it might be the wording that confuses me or the diagram i draw. Could someone who answers it show full working out.



Jason sails from Argos for 12km on a bearing of 245°. He then changes course to sail on a bearing of 335° for 15km.

What is the bearing of Argos from Jason's present position?

Answer to the nearest degree.

 

[Drongoski]

Argos is 116 degrees from Jason's last position.

Unfortunately (and, embarrassingly) I don't know how to post the related diagram, that would have made the solution easy to explain.

 

[Green Yoda]

Argos is 116 degrees from Jason's last position. Unfortunately I don't know how to post the related diagram, that would have made the solution easy to explain.

did you just use geometry or sine/cosine rule?

 

[Green Yoda]

I can't seem to find any angles inside the triangle.

 

[Aeyis]

Argos is 116 degrees from Jason's last position. Unfortunately I don't know how to post the related diagram, that would have made the solution easy to explain.

Definitely the solution would be helpful, is there like a way you can send it via taking a photo on phone or something but whatever, thanks anyways; much appreciated.

 

[Drongoski]

did you just use geometry or sine/cosine rule?

First of all I need to know if my answer is correct.

 

[Drongoski]

Definitely the solution would be helpful, is there like a way you can send it via taking a photo on phone or something but whatever, thanks anyways; much appreciated.

Can you advise if my answer is correct or not? No point providing the solution to a wrong answer.

 

[Aeyis]

Can you advise if my answer is correct or not? No point providing the solution to a wrong answer.

Well, I don't have any answers because this question came from a worksheet with no source on it so yeah.

 

[BLIT2014]

http://www.dance.net/topic/8532810/...gonometry-question-Yr-9-13-hrs.html&replies=2

 

[Green Yoda]

http://www.dance.net/topic/8532810/...gonometry-question-Yr-9-13-hrs.html&replies=2

he used soh cah toa..but how do we know that this is a right angled triangle?

 

[Drongoski]

That diagram is great. Thanks BLIT2014.

You will find that the triangle is right-angled; so there is no need to use the sine or cosine rule! So the left-most angle is the inverse tangent of 12/15 i.e. 0.8. This gives an angle of 38.66... degrees. The angle adjacent and to the to left of this, with the vertical is 25 degrees. So their sum is 64 deg to nearest deg. The supplement of this is 180 - 64 = 116 degrees, and this is the required bearing.

 

[Green Yoda]

That diagram is great. Thanks BLIT2014.

You will find that the triangle is right-angled; so there is no need to use the sine or cosine rule! So the left-most angle is the inverse tangent of 12/15 i.e. 0.8. This gives an angle of 38.66... degrees. The angle adjacent and to the to left of this, with the vertical is 25 degrees. So their sum is 64 deg to nearest deg. The supplement of this is 180 - 64 = 116 degrees, and this is the required bearing.

how do find that this is a right angled triangle?

 

[Drongoski]

he used soh cah toa..but how do we know that this is a right angled triangle?

soh cah toa? That's like riding a bike with training wheels.



Let the 3 vertices, from right to left be A, B and C respectively.

Then AB makes 25 deg with the horizontal thru A, and also 25 deg at B, with the horizontal thru B (alternate angles on // lines)

Now BC makes 25 deg with the vertical thru B; therefore angle ABC is right-angled.