Find the length of PR and the bearing of P from R

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SUMMARY

The discussion focuses on calculating the distance PR and the bearing of point P from point R based on a ship's journey. The ship sails 3 nautical miles from Port P to Port Q at a bearing of 64°, then changes course to 285° and travels 7 nautical miles to Port R. The length PR was determined to be 5.13 nautical miles using the cosine rule, while the bearing of P from R was calculated to be 127.6° after analyzing the angles in the triangle formed by the points.

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Homework Statement


A ship sails 3 nautical miles from Port P on a course of 64° to Port Q, then changes course to 285° and sails a further 7 nautical miles to Port R.
(i) Draw a carefully labelled diagram of the entire route taken. Show the north direction where necessary.
(ii)Calculate the distance PR to three significant figures.
(iii) Determine the bearing of P from R.

Homework Equations

The Attempt at a Solution


I have drawn the diagram which i will post, as you can see I found angle PRQ to be 75° using alternate angles, and I found the exterior angles of the first two bearings by subtracting them from 360°.
But I can't seem to figure out the other two angles in the triangle to find the length, I could simply use the sine rule then°

PLEASE NOTE ON THE DIAGRAM THE ANGLE WHICH SAYS 375° SHOULD BE 285°
 

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Richie Smash said:

Homework Statement


A ship sails 3 nautical miles from Port P on a course of 64° to Port Q, then changes course to 285° and sails a further 7 nautical miles to Port R.
(i) Draw a carefully labelled diagram of the entire route taken. Show the north direction where necessary.
(ii)Calculate the distance PR to three significant figures.
(iii) Determine the bearing of P from R.

Homework Equations

The Attempt at a Solution


I have drawn the diagram which i will post, as you can see I found angle PRQ to be 75° using alternate angles, and I found the exterior angles of the first two bearings by subtracting them from 360°.
But I can't seem to figure out the other two angles in the triangle to find the length, I could simply use the sine rule then
Here is a full size image of your diagram:
bering-png.png

.
Can you determine angle PQR ?
 

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The angle that says 375 should be 285 just to correct that.
I am trying to figure it out... But I Just can't see it, I think it may have to do something with opposite interior angles.
 
Richie Smash said:

Homework Statement


A ship sails 3 nautical miles from Port P on a course of 64° to Port Q, then changes course to 285° and sails a further 7 nautical miles to Port R.
(i) Draw a carefully labelled diagram of the entire route taken. Show the north direction where necessary.
(ii)Calculate the distance PR to three significant figures.
(iii) Determine the bearing of P from R.

Homework Equations

The Attempt at a Solution


I have drawn the diagram which i will post, as you can see I found angle PRQ to be 75° using alternate angles, and I found the exterior angles of the first two bearings by subtracting them from 360°.
But I can't seem to figure out the other two angles in the triangle to find the length, I could simply use the sine rule then
upload_2018-2-2_13-37-51.png
You found the yellow angle as 75° correctly. What is the pink angle then? The angle PRQ is not 75°.
If you know the pink angle, you can calculate the length PR, and then all angles can be determined.
 

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Oh I see Ehild...it makes a corresponding angle... and therefore it would be 180-64+75=41° and I can use the cosine rule... It's actually a bit hard to visualize.. I won't forget this.
 
Richie Smash said:
Oh I see Ehild...it makes a corresponding angle... and therefore it would be 180-(64+75)=41° and I can use the cosine rule... It's actually a bit hard to visualize.. I won't forget this.
Yes, go ahead.
 
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Yes I did it, using the cos rule the length is 5.13, and then for the bearing, if You extend a line from R and call the other point S then the angle NRS is 75° due to the same corresponding angles, so therefore angle NRQ would be 105°.
SO the bearing would be 105°+22.6°(angle PRQ) = 127.6° and it is very correct :) solved thanks a lot guys.
 

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