|
|
|
|
New Member
|
|
May 1, 2009, 04:12 PM
|
|
Mathematics: Topic: Bearings
Please help with this question
I tried working it out, but I couldn't get it at all
2 ships leave port. The first ship sails on a bearing of 270 degrees for 8 km while the second ship sails on a bearing of 200 degrees for 8 km. by drawing, find
(a) the distance between the ship
(b) the bearing of the first ship from the second
(c) the bearing of the second ship from the first
|
|
|
New Member
|
|
May 1, 2009, 04:13 PM
|
|
Maybe try drawing out the problem?
|
|
|
New Member
|
|
May 1, 2009, 04:20 PM
|
|
I did I couldn't get it
|
|
|
New Member
|
|
May 1, 2009, 04:26 PM
|
|
Ahh is there a formula for it? I am trying to think back to Math class but I think we had a formula to figure out this problem.
|
|
|
New Member
|
|
May 1, 2009, 04:27 PM
|
|
c2= a2+b2 - 2(a)(c)Cos x?
|
|
|
New Member
|
|
May 1, 2009, 04:28 PM
|
|
or side a / sin a?
side b/ sin b?
side c/ sin c?
|
|
|
New Member
|
|
May 1, 2009, 04:32 PM
|
|
|
|
|
New Member
|
|
May 1, 2009, 04:33 PM
|
|
Do you have the answers for a and b?
|
|
|
New Member
|
|
May 1, 2009, 04:48 PM
|
|
No I don't :(
|
|
|
New Member
|
|
May 1, 2009, 04:50 PM
|
|
It doesn't help
I need working!
|
|
|
-
|
|
May 1, 2009, 05:32 PM
|
|
Well, this can be solved mathematically, but the question as presented asks for a drawing. It is easiest to do with circular graph paper, but you can simulate that on regular paper if you are very careful in your measurements and use a large enough scale so that the thickness of your lead (might sound funny, but I don't mean it as a joke) doesn't degrade the accuracy of your answer.
If you use a scale something like 2 cm = 1 km you should be OK. Both ships start at the port, which would be the center of the circular graph. Draw one line 16 cm straight west (270 degrees). (or pick a different scale). Draw another line the same length at 200 degrees - which is SSW. The ends of the lines will give you the relative positions of the ships.
Measure the distance between the two positions. Apply your scale and you have the distance between the ships. For the bearing of ship 1 from ship 2, measure the angle offset from North.
With circular graph paper, you can use a parallel ruler lined up on the positions, slide to the center and read the bearing at the upper left part of the circle. On regular paper, there are a number of ways you can do it, but I think the easiest is to drop a perpendicular line from the course of ship 1 such that the line passes through ship 2's position. At the position of ship 2, measure the angle between this line and a line between the ship positions.
Subtract this from 360 degrees to get the bearing. Expect an answer somewhere around 350 degrees.
For the bearing of ship 2 from ship 1, either plot the logical opposite, or simply subtract 180 degrees
|
|
|
New Member
|
|
May 1, 2009, 05:51 PM
|
|
And the distance?
I'm not getting that either
I don't know which angle to use in the cosine rule :(
|
|
|
New Member
|
|
May 2, 2009, 02:59 PM
|
|
Which angle should I use in the cosine rule galactus?
|
|
|
Ultra Member
|
|
May 2, 2009, 03:14 PM
|
|
The law of cosines is used when you have SAS.
You can see from the drawing, the angle needed is 70 degrees?
So, the distance between the ships is:
Also, you can use the law of sines to find the other angles to aid in the other two parts of the problem.
Once you have the above solution, to find the angle at ship A use:
then solve this for A to find that angle.
In other words, solved for angle A, it is
One thing to note is that this is an isosceles triangle, therefore, angle B will be the same as angle A
Once you have those, parts b and c are easy.
|
|
|
New Member
|
|
May 2, 2009, 03:16 PM
|
|
Are you sure galactus?
|
|
|
New Member
|
|
May 2, 2009, 03:16 PM
|
|
How did u get 70 degrees in the first place man?
|
|
|
Ultra Member
|
|
May 2, 2009, 03:24 PM
|
|
How did I get 70 degrees? Come on. What is 270-200?
|
|
|
New Member
|
|
May 2, 2009, 03:30 PM
|
|
Oh Im sorry
I'm not smart like you... so yeah haha
What about the
The bearing of the first ship from the second and
The bearing of the second ship from the first
How do I do that?
|
|
|
New Member
|
|
May 2, 2009, 06:28 PM
|
|
Galactus please help me out!
What about the
The bearing of the first ship from the second and
The bearing of the second ship from the first
How do I do that?
|
|
|
Uber Member
|
|
May 3, 2009, 01:53 AM
|
|
Look more carefully at Galactus' post
Originally Posted by galactus
The law of cosines is used when you have SAS.
You can see from the drawing, the angle needed is 70 degrees?.
So, the distance between the ships is:
Also, you can use the law of sines to find the other angles to aid in the other two parts of the problem.
Once you have the above solution, to find the angle at ship A use:
then solve this for A to find that angle.
In other words, solved for angle A, it is
One thing to note is that this is an isosceles triangle, therefore, angle B will be the same as angle A
Once you have those, parts b and c are easy.
|
|
Question Tools |
Search this Question |
|
|
Add your answer here.
Check out some similar questions!
Mathematics: Topic: Bearings
[ 2 Answers ]
2 ships leave port. The first ship sails on a bearing of 270 degrees for 8 km while the second ship sails on a bearing of 200 degrees for 8 km. By drawing, find:
(a) the distance between the ships
(b) the bearing of the first ship from the second
(c) the bearing of the second ship from the...
Mathematics: Topic: Bearings
[ 2 Answers ]
And what's the bearing Of A from C?
You forgot to answer that man \
Question:
A boat sails 65 km from A to B on a bearing of 140 degrees. From B it sets a new course of 200 degrees towards C, 40 km away. Find by drawing, the distance from C to A( which you answered) and the bearing of A...
Mathematics: Topic: Bearings
[ 2 Answers ]
How did you get 50 degrees and 20 degrees and all that
And weren't you also suppose to find the distance in cosine rule
Like subract from a certain degree and then use because rule?
I want the working for that
Mathematics: Topic: Bearings
[ 2 Answers ]
A boat sails 65 km from A to B on a bearing of 140 degrees. From B it sets a new course of 200 degrees towards C, 40 km away. Find by drawing, the distance from C to A and the bearing of A to C
Answer:
View more questions
Search
|