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Alex-parker · May 1, 2009, 06:01 PM

I have something similar to Lee
but I don't understand it
explain it please

A car drives from A 280 km on a bearing of 075° to B. From B its course is 012° for a further 120 km to c.

1. how far B is north of A
2. how far is C north of A
3. how far c is east of A

please respond A.S.A.P

thank you very much.

Alex= sexxy boy

Perito · May 1, 2009, 07:31 PM

I won't do it for you. That wouldn't help you learn and unless you're learning this, what good is it to study it? I will, however, tell you how it's done.

Draw this on graph paper. 0 degrees (and 360 degrees) is due north. 90 degrees is east (I think). 180 degrees is south, and 270 degrees is west.

Draw right triangles from obvious points on the graph. For example, from the origin to point B. This is a right triangle with angles 75 degrees, 15 degrees. You know the length of O (origin) B, and the angles so you can solve the entire triangle using trigonometry.

From B, draw another triangle and work that out.

Zazonker · May 2, 2009, 12:11 AM

A little clarification. Perito has it right. His is a relatively straight forward way to solve the problem via calculation (rather than straight graphing.

To help visualize a little, from the origin headed 075 - the 15 degree angle is measured from east counter-clockwise. Running a perpendicular from east to point B completes your right angle. From B to C, the 012 angle is measured from North clockwise. Once you solve for the sides of the two triangles you can see which sides you need to add together to get the answers to the questions.

The driver of the car is likely to be arrested since I rather doubt that there are any actual roads in the world matching this problem.

Alex-parker · May 2, 2009, 09:44 AM

Hehe dudes
I still don't understand

galactus · May 2, 2009, 10:03 AM

Let's try implementing a coordinate system.

Let the origin, where the car starts, be (0,0). That is, the north coordinate is 0 and the East coordinate is 0.

Just make a diagram and you can see from A to B we have a right triangle with angle A being 15 degrees.

$East=x=280cos(15)$

$North=y=280sin(15)$

or

$East=x=280sin(75)$

$North=y=280cos(75)$

Now, add this to the previous coordinates and we find that the coordinates of B are

x=270.46, y=72.47

Continue and try the other from B to C.

That inner angle at B is 117 degrees (see why?).

Now use the law of cosines to find the distance from A to C.

Unknown008 · May 2, 2009, 10:17 AM

Can you see it now, that you've been given the drawing?

Alex-parker · May 2, 2009, 10:24 AM

How did u get
280 because 75 and all that
Man I'm so confused

Unknown008 · May 2, 2009, 10:28 AM

I think that galactus gave you answers that are too far for you to understand now, perhaps in a year or two you'll understand.

Then it's not 280 cos 75 but 280 cos 15 OR 280 sin 75! :p

Alex-parker · May 2, 2009, 10:41 AM

I know, exactly, I didn't understand at all
What are we FINDING over here!
Please help me haha

Because it asks to answer
1. how far B is north of A
2. how far is C north of A
3. how far c is east of A

And don't know what I have to do
Because my teacher never taught me this

Zazonker · May 2, 2009, 10:46 AM

I am a little confused about your level of understanding. Perhaps you can help us help you by answering a few questions.

1. Do you understand the circular coordinate system (North = 0 or 360 degrees, East = 90 degrees, South = 180 degrees, West = 270 degrees) and values in between any two of these refer to directions in between them.

2. Do you understand what a bearing is?

3. Do you know how to solve a right triangle?

4. Do you know the definitions of sine and cosine?

5. What grade are you in?

Unknown008 · May 2, 2009, 10:52 AM

1. Consider only the distance from A to B, in the north direction.

Well, that's my guess... I just looked your question, without paying much attention to the 1. 2. and 3. :o

I'll do the first one for you. The vertical distance from A to B is 280 sin (x), if you draw a right angled triangle, like the one below. X is 90 degrees minus 75 degrees.

Unknown008 · May 2, 2009, 10:54 AM

The vertical distance from the right angle to B is what you're probably been asked to look for.

Since sin x = [your unknown] / 280,

280 sin (x) = [your unknown]

Alex-parker · May 2, 2009, 12:59 PM

So how far is B north of A?

galactus · May 2, 2009, 01:13 PM

I gave you that one and so did Unknown: 280sin(15)

Or 280cos(75)

Alex-parker · May 2, 2009, 01:28 PM

Ooooo
Thanks :)

But hey
Then how do I find how far is C north of A then?

Alex-parker · May 2, 2009, 01:30 PM

3. And C east of A

P.S
Sorry I'm blindly asking these questions man
But my teacher didn't explain me anything like this... I swear
And I'm confused

Alex-parker · May 2, 2009, 03:39 PM

mr. galactus

How do I find?
2. how far is C north of A
3. how far c is east of A

Unknown008 · May 3, 2009, 01:35 AM

Gosh! I did one for you, and so did galactus, and you still can't figure out how to solve that? I've done the first one, apply the same method and you'll have the answer. Post your answers. We'll see if you did it right, but try it sincerely. We are here to HELP, not GIVE ANSWERS!