Who wants to help me with some work?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Who wants to help me with some work?

Postby Sea Ferring » Mon Dec 22, 2008 1:04 am UTC

I take math by correspondence (home schooling pretty much) for personal reasons and am doing trigonometry at the moment and would really appreciate if someone helped me with this question:

Andrea, Fred, and Gurtek leave a dock and sail 10 km at a bearing of 090° to a fishing spot. After fishing for a while, they sail on a bearing of 210° to another spot 6km away. the boat travels at 15 km/h of the time is now 3:30 PM, how long can they continue to fish at this location if they must return to the dock before dark at 6:30 Calculate your answer to the nearest minute.

Diagrams and explanations please. :)

I don't really have a teacher to help me here (Christmas break) and the textbooks are of little help for this question, I don't necessary need the answer just a labeled diagram and a hint. Please.
Image

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Mon Dec 22, 2008 2:28 am UTC

The first thing you should do is translate bearings into directions. A bearing of x° is a direction which is x° clockwise of due north, so a bearing of 0° is due north, 90° is due east, 180° is due south, and 270° is due west. Once you've done the translation, you should be able to diagram it yourself, and once you have the diagram hopefully you can get it without further hints.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
Yakk
Poster with most posts but no title.
Posts: 11120
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove

Re: Who wants to help me with some work?

Postby Yakk » Mon Dec 22, 2008 3:13 pm UTC

Work out where they are, relative to their starting location.

I'd (personally) assume they are going to go strait from their current location to the dock. Work out how long that will take.

Now work out how long they have to wait.

Do that, and use skeptical's plan to draw the diagram. Does it fall out? (Remember, coming back to explain your answer, if you got it, would be polite!)
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Mon Dec 22, 2008 5:29 pm UTC

Yakk wrote:(Remember, coming back to explain your answer, if you got it, would be polite!)

Or at least let us know you got it. The only reason to explain what you've done would be if you want us to check it for mistakes, or if you get stuck again and want some more help.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
z4lis
Posts: 767
Joined: Mon Mar 03, 2008 10:59 pm UTC

Re: Who wants to help me with some work?

Postby z4lis » Mon Dec 22, 2008 6:23 pm UTC

If you know anything about vectors, turn all of their movements into vectors and add. Then solve the problem.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Mon Dec 22, 2008 10:15 pm UTC

I think it's just basic trig stuff so 210 degrees is equal to -30 degrees and since we have 90 degrees and triangles always add up to 180 degrees the missing degree must be 60 degrees (- = +)

I think, am I in the right direction?

My mistake
Image

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Mon Dec 22, 2008 10:47 pm UTC

I think I may of messed up somewhere, if someone could show me what this should look like (a drawing) it'd be great, please excuse the writing and eraser marks, this is not my forte.

http://www.vulomedia.com/images/702931222200834549PM.jpg
Image

Seraph
Posts: 342
Joined: Mon Jul 16, 2007 4:51 pm UTC

Re: Who wants to help me with some work?

Postby Seraph » Mon Dec 22, 2008 11:02 pm UTC

Sea Ferring wrote:I think I may of messed up somewhere, if someone could show me what this should look like (a drawing) it'd be great, please excuse the writing and eraser marks, this is not my forte.

http://www.vulomedia.com/images/702931222200834549PM.jpg

Your triangle is cleary wrong.
Side AB is labled at 6km, and side AC is labeled as 10km, but when you look at the angles side should be AB the longest side of the triangle because it's opposite the biggest angle.

Also, your angle A doesn't match the problem description for the angle between the 6 km and 10 km legs.

There is also no indication in the problem that the triangle formed by the ships trip will be a right triangle. My guess is that this is a Law of cosines problem.

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 3:10 am UTC

Seraph wrote:
Sea Ferring wrote:I think I may of messed up somewhere, if someone could show me what this should look like (a drawing) it'd be great, please excuse the writing and eraser marks, this is not my forte.

http://www.vulomedia.com/images/702931222200834549PM.jpg

Your triangle is cleary wrong.
Side AB is labled at 6km, and side AC is labeled as 10km, but when you look at the angles side should be AB the longest side of the triangle because it's opposite the biggest angle.

Also, your angle A doesn't match the problem description for the angle between the 6 km and 10 km legs.

There is also no indication in the problem that the triangle formed by the ships trip will be a right triangle. My guess is that this is a Law of cosines problem.

But it says they travel 90 degrees thus it should be a right triangle shouldn't it?

Wait, it should look more like this, I am over thinking this aren't I? :oops: http://www.vulomedia.com/images/4567Picture17.jpg

I also have another question that I'll post after this.
Image

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 3:52 am UTC

Alright I think it's 5.7 km to dock

Now how do you figure out how long it will take to go 5.7 km if you are traveling at a constant of 15 km/h?
Image

User avatar
Yakk
Poster with most posts but no title.
Posts: 11120
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove

Re: Who wants to help me with some work?

Postby Yakk » Tue Dec 23, 2008 3:55 am UTC

Ah, so you don't understand what is going on at all.

The 'bearing 90 degrees' is a direction. Both the 90 degrees and 210 degrees are directions relative to "due north".

Have you ever done any orienteering?

90 degrees means "90 degrees clockwise of north" -- or , due east.

210 degrees ... well, that's 30+180 degrees. 180 degrees is due south, +30 degrees, so you get 30 degrees clockwise of due south.

Nothing in this implies that there will be a right angle triangle on the route they travel.

You seem to be pattern matching -- you see the number 90 and think "this is a right angle triangle problem". That about explain why you thought it should have a right angle in the solution?

Sea Ferring wrote:Alright I think it's 5.7 km to dock

Now how do you figure out how long it will take to go 5.7 km if you are traveling at a constant of 15 km/h?


In order of increasing difficulty, some questions that will lead you to the answer I hope:

How long does it take to travel 10 km if you travel at a constant 10 km/h?

How long does it take to travel 5 km if you travel at a constant 10 km/h?

How long does it take to travel 4.3 km if you travel at a constant 10 km/h?

How long does it take to travel 15 km if you travel at a constant 15 km/h?

How long does it take to travel 10 km if you travel at a constant 15 km/h?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 4:04 am UTC

Yakk wrote:Ah, so you don't understand what is going on at all.

The 'bearing 90 degrees' is a direction. Both the 90 degrees and 210 degrees are directions relative to "due north".

Have you ever done any orienteering?

90 degrees means "90 degrees clockwise of north" -- or , due east.

210 degrees ... well, that's 30+180 degrees. 180 degrees is due south, +30 degrees, so you get 30 degrees clockwise of due south.

Nothing in this implies that there will be a right angle triangle on the route they travel.

You seem to be pattern matching -- you see the number 90 and think "this is a right angle triangle problem". That about explain why you thought it should have a right angle in the solution?

Sea Ferring wrote:Alright I think it's 5.7 km to dock

Now how do you figure out how long it will take to go 5.7 km if you are traveling at a constant of 15 km/h?


In order of increasing difficulty, some questions that will lead you to the answer I hope:

How long does it take to travel 10 km if you travel at a constant 10 km/h?

How long does it take to travel 5 km if you travel at a constant 10 km/h?

How long does it take to travel 4.3 km if you travel at a constant 10 km/h?

How long does it take to travel 15 km if you travel at a constant 15 km/h?

How long does it take to travel 10 km if you travel at a constant 15 km/h?


I know but 15/ 5.7 = 2.6315

How is that in terms of minutes?
Image

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 6:04 am UTC

Solved (I hope).
Image

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Tue Dec 23, 2008 7:07 am UTC

Sea Ferring wrote:Alright I think it's 5.7 km to dock

Now how do you figure out how long it will take to go 5.7 km if you are traveling at a constant of 15 km/h?

FYI, unless I'm making a mistake in mental math, it's not 5.7 km to dock.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 7:14 am UTC

skeptical scientist wrote:
Sea Ferring wrote:Alright I think it's 5.7 km to dock

Now how do you figure out how long it will take to go 5.7 km if you are traveling at a constant of 15 km/h?

FYI, unless I'm making a mistake in mental math, it's not 5.7 km to dock.


b²= a²+c²- 2ac cosB
b²= 6²+10²- 2*6*10Cos(210)
b²= 136- 104
b= 32 square rooted
b= 5.656854248 rounded to 5.7

Where'd I go wrong?
Image

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Tue Dec 23, 2008 7:23 am UTC

210 is not an interior angle in any triangle, and certainly not the angle you want here.

You seem to still not be understanding the meaning of "bearing". I recommend drawing a picture, first indicating the direction of north, then draw an arrow pointing in the direction indicated by a bearing of 90° (which has now been stated twice), and another arrow in the direction indicated by a bearing of 210°. Then label the dock, draw a line of about 10cm in the direction of the first arrow, and then starting from the end of that line draw another line of about 6cm in the direction of the second arrow.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 7:32 am UTC

skeptical scientist wrote:210 is not an interior angle in any triangle, and certainly not the angle you want here.

You seem to still not be understanding the meaning of "bearing". I recommend drawing a picture, first indicating the direction of north, then draw an arrow pointing in the direction indicated by a bearing of 90° (which has now been stated twice), and another arrow in the direction indicated by a bearing of 210°. Then label the dock, draw a line of about 10cm in the direction of the first arrow, and then starting from the end of that line draw another line of about 6cm in the direction of the second arrow.


Could you perchance draw it?
Image

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Tue Dec 23, 2008 8:33 am UTC

Sorry, I'm at home on my laptop with no good drawing program and no scanner. Seriously though, it's not that hard to figure out what a bearing is, and that's the only thing you need to solve the problem.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
Sea Ferring
Posts: 53
Joined: Wed Nov 12, 2008 3:55 am UTC
Contact:

Re: Who wants to help me with some work?

Postby Sea Ferring » Tue Dec 23, 2008 7:30 pm UTC

skeptical scientist wrote:Sorry, I'm at home on my laptop with no good drawing program and no scanner. Seriously though, it's not that hard to figure out what a bearing is, and that's the only thing you need to solve the problem.


Please tell me this is right, I started over for neatness.

http://www.vulomedia.com/images/23549Picture18.jpg
Image

User avatar
skeptical scientist
closed-minded spiritualist
Posts: 6142
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

Re: Who wants to help me with some work?

Postby skeptical scientist » Tue Dec 23, 2008 7:55 pm UTC

No, it's not right. 210 is not the angle between the two directions, it's the angle between the second direction and due north, while 90 is the angle between the first direction and due north.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

User avatar
z4lis
Posts: 767
Joined: Mon Mar 03, 2008 10:59 pm UTC

Re: Who wants to help me with some work?

Postby z4lis » Tue Dec 23, 2008 8:05 pm UTC

No, you've got the 210 degrees as an angle between the two paths the boat takes. That's not what a bearing is. To use a bearing, you choose a line, say the x-axis. (Note: you could replace every instance of "x-axis" in here with "north" and it'd be the same. EDIT: OK, not quite the same. The directions would change, but this parenthetical statement is largely irrelevant anyway.) Then if you go a direction with a bearing of x degrees, then the line you travel along intersects the x-axis at x degrees, not the line along which you previously traveled.

For example, if the x-axis points east and I travel with a bearing of 90 degrees, I'm going north. No matter what path I've just taken. I might sail east, west, anywhere. But then if I stop and travel with a bearing of 90 degrees, I'm always going north. If my bearing is 180 degrees, I'm going west. 270, and I'm going south. 45 degrees would be northeast.

So for your problem, the boat will travel some distance (I forgot the value) north, then travel some distance in a direction that's roughly southwest. To see it, draw a horizontal (parallel to the x-axis) line through the place the boat is after the first trip. Then imagine a little arrow pointing to the right along that line, and rotate it 210 degrees counter-clockwise. That's the direction the boat goes in.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 17 guests