Components of a vector - don't know what I am doing wrong

[Hemingway]

Homework Statement

Two medical field teams set out for a reconnaissance mission and lose one another. The GPS at their original camp shows the location of the first medical team as 43 km away, 15° north of west. The second medical team strayed 32 km away, 33° east of north from base camp.

When the first team uses its emergency GPS to check the position of the second team, what does it give for the second team's (a) distance from them and (b) direction, measured from due east?

Homework Equations

Rx = ax + bx
Ry = ay + by
a² + b² = c²
tan ^-1 = opp/adj

The Attempt at a Solution

ax = cos 15° x 43km
= -41.53km

ay = sin 15° x 43km
= 11.13km

bx = sin 33° x 32km
=17.42

by = cos 33° x 32km
= 26.83

Rx = ax + bx
= -41.53km + 17.42km
= - 24.11km
(or if we take positive direction of x the other way)
= +24.11km

Ry = ay +by
= 11.13km + 26.83km
= 37.96km

Rx² + Ry² = R_hypotenuse
= [tex]\sqrt{}-24.11² + 37.96²[/tex]
= $$\sqrt{}2022.3$$
= 44.97km

tan^-1 = opp/adj
= 57.6°

both answers were wrong. I cannot see what little mistake I am making

 

[Hemingway]

Sorry about code problem, hopefully you understand what I mean. ignore the 'sup'
=√( 24.11 ^2 + 37.96^2)
= √(2022.3)
= 44.97km

 

[diazona]

Yeah, it's clear. You're just mixing up the LaTeX and BBcode syntaxes but we're kind of used to both around here

Rx = ax + bx
Ry = ay + by

Take a close look at those formulas. They're not correct. (Hint: if a and b are the same vector, what should you get for R?)

 

[Hemingway]

Okay so I worked out that Rx = ax + bx but ignored the Ry as it wasn't required, turned out to be correct. So thank you for that.

My only drama at this point is the angle for east. I apply an axis for direction at team A and look for angle at east. I wish I knew how to diagram this on here. But basically incorrectly assumed the east angle is the angle at the left vertice of the triangle, And since the I am not sure if this is a correct assumption. I struggle with direction sometimes. There is no right angle so I cannot do that, so need geometry rules am I correct-ish??

 

[Hemingway]

Sorry that was a brain dump. I cannot use inverse tan /cos/sine here as it is not a right angled triangle correct? and is my assumption of measuring the angle from team A's corner of triangle correct? Thank you so much by the way!

 

[diazona]

Okay so I worked out that Rx = ax + bx but ignored the Ry as it wasn't required, turned out to be correct. So thank you for that.

Did you notice I said the formula Rx = ax + bx was incorrect? If you got the right answer using that formula, it was completely by accident.

My only drama at this point is the angle for east. I apply an axis for direction at team A and look for angle at east. I wish I knew how to diagram this on here. But basically incorrectly assumed the east angle is the angle at the left vertice of the triangle, And since the I am not sure if this is a correct assumption. I struggle with direction sometimes. There is no right angle so I cannot do that, so need geometry rules am I correct-ish??

But you can draw a right triangle

If you can draw a picture of what you're talking about using some image editor, you can attach it to your next post. I think that might help.

 

[Hemingway]

Did you notice I said the formula Rx = ax + bx was incorrect? If you got the right answer using that formula, it was completely by accident.

But you can draw a right triangle

If you can draw a picture of what you're talking about using some image editor, you can attach it to your next post. I think that might help.

Ah, damn it. I did see that but obviously I was misled by the fortuitous accident :) I got the angle wrong obviously so I have been given a new set of numbers now for the problem. I will post my working, images and ideas a little later after I go to work.

Thank you so much for your assistance diazona, I really very much appreciate you taking the time and patience :)

H.