Bearing and Distance

[Chikis]

Town Q is 20km due north of P. The bearing of town R from Q is [latex]140^0[/latex]. If R is 8 km from Q. calculate:

 

(a) the bearing of R from P, to the nearest degree:

 

(b) how far north of P, R is, correct to 2 significant figures;

Drawing the diagram is my main problem. If you can help me with the drawing, I can attempt the working.

Edited April 22, 2014 by Chikis

[studiot]

Bearings are measured clockwise from the vertical axis on paper. This is the north axis on the ground and on the paper.

 

I have started your sketch off by drawing a north axis from P due north and positioning Q on it 20km north.

 

Then I have drawn a line from Q at 140 clockwise from this north axis towards R.

 

Then I marked R on this line (which is the bearing of R from Q) at a distance of 8km.

 

At this point I will leave you to carry on with the diagram to draw in the three extra lines needed to complete both parts of your problem, which is then just an exercise in trigonometry.

 

Post your results when you have done this.

 

[Chikis]

Thank you for the help. But I still have problem completing the drawing to show the bearing of R from P. Please can you complete it for me or just describe how I can do it?

[studiot]

 

Bearings are measured clockwise from the vertical axis on paper. This is the north axis on the ground and on the paper.

 

I may not do your homework for you, only help you do it for yourself.

 

Do you understand this statement I made in post #2

 

You have an example bearing drawn in my sketch.

 

Do you have a North line through P?

Edited June 16, 2014 by studiot

[daniton]

Hello Guys,

I've been away for a while for some reason. It's good to be back. Anyway, Chikis I understand from your question that you just have to find the inner most angles and the remaining side of the triangle formed by point P,Q and R using the Trigonometric Laws( Cosine and Sine Laws as your wish). I hope you know them. Studiot explained you very briefly what bearing is. You just have to interpret the angles you find to bearings.

[Chikis]

I have managed to complete the diagram. The diagram has the description:

triangle QPR. <QPR = [latex]140^0[/latex]. QP = 20km and PR = 8km.

 

I first used cosine rule to find PR, which I got as 14.79km. I again used cosine rule to find <PQR - bearing of R from P, which I got as [latex]20^0[/latex] to the nearest degree. Now, how do you see my work?

 

As for the second part of the question (b), am yet to understand it fully. How do they mean by "how far north of P, R is". From the way I look at it, in the diagram Q is 20km north of P. Or do they mean the distance PR, which I got as 14.79km?

Edited June 17, 2014 by Chikis

[studiot]

Cosine rule?

Wow you really are doing things the difficult way. I also think you have a typo in your posted angle QPR.

 

 

triangle QPR. <QPR =

Perhaps you are not comfortable using bearings?

 

I have prepared an 8 step guide to completing this problem.

 

1) Position P

 

2) Draw a North line through P

 

3) Mark on Q, 20km north of P. Note the North line through Q is also the same as the North line through P, since Q is due North of P.

 

4) At Q mark out a line at 140 deg from the North Line at Q.

 

5) Mark R 8km along this line. Note also that smaller angle PQR = (180 - 40) = 40

 

6) Join PR to form triangle PQR, Call the angle QPR theta. This angle is the bearing of R from P because it is the clockwise angle of PR from the north line PQ, at P.

 

7) From R draw line RS at right angles to PQ, meeting PQ at S. You now have two smaller right angled triangles and in triangle SQR the hypotenuse is 8.

Calculate SR = 8sin40

Calculate QS = 8cos40

 

8) S is the same distance north of P as R and PQ= 20 = PS + QS

So calculate the distance R is north of P = SP = QP-QS

 

tan (Theta) in triangle SRP is SR/SP so calculate the bearing of R from P = theta = tan^-1(SR/SP)

Edited June 17, 2014 by studiot