Calculate Displacement Between Two Oases Without a Coordinate System

[TeenieWeenie]

Homework Statement

Starting from one oasis, a camel walks 82.021 ft in a direction 30 degrees south of west and then walks 30 km toward the north to a second oasis. Without using a coordinate system, calculate the magnitude and the direction of the displacement from the first oasis to the second.

Homework Equations

I don't know what to use, nothing comes to mind.

The Attempt at a Solution

No attempt, I don't get what to do.
I can kinda do it with a coordinate system...but I can't do it without.

 

[Thaakisfox]

Draw the vector diagram. The displacement vector is the vector between the beginning and the endpoint. Use the cosine law to find its magnitude and basic geometry to get the angle.

 

[TeenieWeenie]

Okie.

I end up with origin pointing 30 degrees west of north (120 degrees)... (edit:30 degrees north of west?)
What is the cosine law? I'm having trouble finding it in my textbook.
I'm still lost.

 

[Thaakisfox]

Well that doesn't seem quite right. how did you calculate that?

 

[TeenieWeenie]

|\
------
|/

-- = origin

|/ = 30 degrees south of west = 60 degree angle?

|\ = 30 degrees north of west = 60 degree angle? = 120 degrees?

I'm super lost with this problem :(
I think one of the numbers is wrong (82.021 ft vs 30 km is a ginormous difference) but I can ignore that for now. I just need to understand how to complete the problem :(
cosine law: Avec dot Bvec = ABcosO = |Avec||Bvec|cosO ?

where Avec = (82.021, cos30 degrees) ?
Bvec = (30km, cos60 degrees)?

 

[Thaakisfox]

Probably the first number is 82021 i.e. a bit more than 82000 feet. Then that is about exactly 25km.

Yes, so consider the direction west to be the x-axis. Then the camel walk on a line which is 30 degree angle with the x axis. After it walked 25km, it turns north and walks another 30km.
No, that's not the cosine law.

 

[TeenieWeenie]

That might be so. :Z

Yes, the camel starts off walking in the south west direction (total angle of 210 degrees).
After the 25km walk, it turns 60 degrees to face north and walks 30km?

Hmm, what is the cosine law :?
a^2 = b^2 + c^2 - 2bc*cos\alpha\
b^2 = a^2 + c^2 - 2ac*cos\beta\
?

 

[Thaakisfox]

Yes that's it.

you are right that's the cosine law. You know the b (25km) you know the c (30km) you know the angle between them (60 degrees) so from here you can find the displacement (a)

 

[TeenieWeenie]

Do I assume that the 30km walk north completes a 90 degree right triangle?
So logically, the camel could have went straight west and hit the 2nd oasis?

But then wouldn't the hypotenuse = 25km, which is less than the base = 30km?

 

[Thaakisfox]

No you don't assume that it is a right angle.

 

[TeenieWeenie]

Ooo yeah...which is why we use law of cos!

a²=25²+30²-[(2*25*30)cos60^o]

a²=625+900-[(2*25*30)cos60^o]
a²=1525 -1500cos(60^o) = 775
a=5*(31)^1/2

so that's the magnitude of the a vector?
now direction would be 30^o north of west with a distance of 5*(31)^1/2km?

How would I answer the question properly?

 

[Thaakisfox]

Yes that's the magnitude. But no the direction is a bit more tricky. Use the sine and cosine function to express that angle. (i.e. you have the three sides of a triangle and you know one of its angles is 60, now use sines and cosines to figure out the other angles.)

 

[TeenieWeenie]

I know my angle A is 60 degrees because the camel walked 30 degrees south of west and then turned straight ahead north, thus creating a 60 degree turn.
Correct me I'm wrong please -_-

using other law of cos:
solving for B angle:
25²= (5(31)^1/2)² + 30² - 2 (5(31)^1/2(30)cosB

625 = 775 + 900 -2(5(31)^1/2(30)cosB
-1050 / (-2(5(31)^1/2(30)) = cosB
according to my calculations: B is almost 90 degrees (it's actually 89.978 degrees).

solving for A angle:
900 = 775 + 625 - 2(5)(root31)(25)cos C
-500 = - 2(5)(root31)(25)cos C
2 = (root31)cosC
C = ~69 degrees.

How can this be so? 90 + 69 + 60 = 219 degrees > 180!

 

[Thaakisfox]

calculate it again ;) that's not 90 degrees.