J: I'm in a pickle :L

D

Denzel

New member
Joined
Mar 13, 2019
Messages
45
I honestly do not know where to start, I was given a hint to use Rcosθ but..

Math.png
 
Last edited by a moderator:
Another hint: the latitude 37 degrees is the angle from the equator to the circle. What is the angle from the north pole? Add in to the diagram the line joining the north pole to the center, and the radius to a point on the circle. Then think about how to find the radius of the circle itself.
 
Dr.Peterson said:
Another hint: the latitude 37 degrees is the angle from the equator to the circle. What is the angle from the north pole? Add in to the diagram the line joining the north pole to the center, and the radius to a point on the circle. Then think about how to find the radius of the circle itself.
Click to expand...
37 degrees from the north pole? Sorry. I don't understand what you mean. :/
 
And that's what I said. So why would you say that the angle from the north pole is 37° ?

Here's a picture:

FMH115090.png
 
I would just use the latitude given, and some basic trig to find the radius \(a\) of the circle:

mhb_0009.png

How can we relate the 3 quantities?
 
Okay, so, what I did was find the circumference of the earth 2πr and the multiply it by cos 37°. This was question 1 ( And I still don't fully understand it though.. :l )
 
Denzel said:
Okay, so, what I did was find the circumference of the earth 2πr and the multiply it by cos 37°. This was question 1 ( And I still don't fully understand it though.. :l )
Click to expand...

That will work, but what I was asking you to do is use the definition of the cosine function on the right triangle in my diagram to state:

[MATH]\cos(\theta)=\frac{a}{r_E}\implies a=r_E\cos(\theta)[/MATH]
And so the length or circumference \(C\) of the circle is:

[MATH]C=2\pi a=2\pi r_E\cos(\theta)[/MATH]
This is what you did, but it shows why it works. :)
 
MarkFL said:
That will work, but what I was asking you to do is use the definition of the cosine function on the right triangle in my diagram to state:

[MATH]\cos(\theta)=\frac{a}{r_E}\implies a=r_E\cos(\theta)[/MATH]
And so the length or circumference \(C\) of the circle is:

[MATH]C=2\pi a=2\pi r_E\cos(\theta)[/MATH]
This is what you did, but it shows why it works. :)
Click to expand...
I see.
 
For question (ii), we could use the arc-length formula:

[MATH]s=r\theta\implies \theta=\frac{s}{r}[/MATH]
We are given \(s\), and \(r=r_E\cos\left(37^{\circ}\right)\). We need to express \(\theta\) as a function of \(x\), by computing the difference in the given longitudes, and this angular difference needs to be in radians. Can you proceed?
 
I tried but I give up now, after watching a lot of videos, I tried - 31,948.3 - 5390 = 26, 558.3. Then I used θ= s/r . I feel stupid (._. )
 
We have one location that is \(50^{\circ}\) west of the prime meridian, and another that is \(x^{\circ}\) east of the prime meridian, hence:

[MATH]\theta=(50+x)^{\circ}\frac{\pi}{180^{\circ}}=\frac{(50+x)\pi}{180}[/MATH]
Thus we want:

[MATH]\frac{(50+x)\pi}{180}=\frac{5390}{6370\cos\left(37^{\circ}\right)}[/MATH]
Can you solve for \(x\)?
 
Thank you, Thank you, SO much! It was my first time doing this( Longitudes and Latitudes, arc length), I've never seen something like this before. You guys were a great help. I'll have to do some more.
1553277764254.png
 
Yes, good! I get:

[MATH]x\approx10.70484419169697^{\circ}\quad\checkmark[/MATH]
 
You must log in or register to reply here.
Top