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not physics

Vector addition?

application of vector addition, yah.

i somehow figured out how to get resultant, but not the angle...

im not suppose to use component method

Use tan(y/x)

Oh I see. THen I'm afraid I can be of no help to you.

ok...oh well. thx u

draw it out, and remember when directions are concerned, 0 degrees is North.

yeah, i forgot it was bearings

O.K, good luck on this problem it is a good question

|dw:1333520833003:dw|

ur question is a hard 1. i tried 8 times and cant figure it out. sorry

ok. its alright.

Okay ive done this before. Let me dig up my notes.

hello?

computer went wonky.

Sorry. Were you given the answer?

yeah.its 154.0km/h at a bearing of 211.9 degrees.

Draw your version of the diagram. I wanna see if we are using the same one

i drew 2...

or do you want what the txtbook has>?

yeah that would be good

|dw:1333522792176:dw|

I like how "not physics" is the first thing NotTim says.

you know, just in case.

So we are finding the resultant vector?

i already found it (somehow...I don't really know how). I'm looking for the angle now.

bump

argghh. that's component method. im not suppose to use component method!

since they don't have same directions, i don't think you have other choice ...

what about application of vector additions? using sin adn cos law?

let's try it.

\[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\]
subtract from 225

r^2=150^2+35^2-2(150)(35)cos90

theta is angle between resultant and original flight

wait...so trig ratios work, even with non-right angle triangles?

|dw:1333520171086:dw|

|dw:1333520250197:dw|

225-135=90

its 315...

im sure thats not an error in my grammar

blowing from a bearing of 315º is the same as blowing at a bearing of 135º

|dw:1333520364868:dw|

dock, i dont udnerstand. how would i derive that info?

i think that would be 180 - (315-225)
and apply cosine law ..
i think that might work

if i tell you wind is coming from in front of you, which way is it going to push you?

the same way its coming from?

you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector

yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?

|dw:1333520788256:dw|
you could do as you did above.

draw a line from 315 bearing through the origin

its just intuitive, not really something you should have to prove

just the wording of the problem

no, but how do i figure that out myself? to get 135 from the equilibrant of 315?

i dont know if im making sense here...

they aren't giving you the true bearing of the wind's velocity

they are giving it to you from a different reference

its so far up now. what angle did i derive?

135

the wind blowing.

How did you calculate 135degrees?

315-180

oohhhkay.

the observer (plane) and wind, if looking at each other, form a 180º line of sight

or a straight line

ok.

now that i figured that out, what's our next step?

we werent taught arctan (to add to my problems)

or, that was long before

this will give you the angle between the resultant and the plane's direction

oh. trig ratios. tanangle=o/a right

arctan(35/150) is approximately 13.13º

how do i know which one is my opposite and adjacent?

|dw:1333521664877:dw|

oh bugger. I thought we were looking for another angle. this clarfies much,.

as you can see, the magnitutde can also be calculated with pythagorean theorem

yes.

i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.

thx with sticking with me here folks, greatly appreciated.

subtract that angle from the plane's bearing

yw