# Math Help - Trigonometry word problem

1. ## Trigonometry word problem

I've tried every which way to think about this problem but i'm unsure as to how to start and which way to go with it.

New York and Los Angeles are 2450 miles apart. Find the angle that the arc between these 2 cities subtends at the center of the earth given the radius of the earth is 3960.

2. Originally Posted by JonathanEyoon
I've tried every which way to think about this problem but i'm unsure as to how to start and which way to go with it.

New York and Los Angeles are 2450 miles apart. Find the angle that the arc between these 2 cities subtends at the center of the earth given the radius of the earth is 3960.
The Earth is a sphere. That means that if we follow any "great circle" (which is how we figure out the "straight line" distance on the surface of the Earth) we are following a portion of a circle.

There is a relationship between the arc length, radius, and central angle of a wedge of a circle: $s = r \theta$, where $\theta$ is the angle in radians.

See if you can use this.

-Dan

3. Originally Posted by topsquark
The Earth is a sphere. That means that if we follow any "great circle" (which is how we figure out the "straight line" distance on the surface of the Earth) we are following a portion of a circle.

There is a relationship between the arc length, radius, and central angle of a wedge of a circle: $s = r \theta$, where $\theta$ is the angle in radians.

See if you can use this.

-Dan

I know i'm supposed to use either the law of sines or law of cosines to solve this. MmM... I thought about what you wrote and i'm still clueless

4. Originally Posted by JonathanEyoon
I know i'm supposed to use either the law of sines or law of cosines to solve this. MmM... I thought about what you wrote and i'm still clueless
topsquark mentioned only circles to you, how is it that you came up with rules that have to do with triangles?

the arclength is 2450, the radius is 3960, all you need to do is find $\theta$ and topsquark was kind enough to give you the formula for that

5. Originally Posted by Jhevon
topsquark mentioned only circles to you, how is it that you came up with rules that have to do with triangles?

the arclength is 2450, the radius is 3960, all you need to do is find $\theta$ and topsquark was kind enough to give you the formula for that

I came up with the rules to find triangles because that's the unit we're on right now and the problem is from the chapter review. So i kinda naturally assumed it had to make use of the Laws of Sine or Cosine . Is there anyway you could draw a diagram for me. I'm just having the worst time picturing what it is I am supposed to look for.

6. Originally Posted by JonathanEyoon
I came up with the rules to find triangles because that's the unit we're on right now and the problem is from the chapter review. So i kinda naturally assumed it had to make use of the Laws of Sine or Cosine . Is there anyway you could draw a diagram for me. I'm just having the worst time picturing what it is I am supposed to look for.
your diagram is fine. but you do not have a triangle here. what you have is called a sector. notice that one side of the triangle is not straight, the sine and cosine rules do not apply because you don't have a triangle. use topsquark's formula

(if you really want to, you can use the cosine rule, but it is overkill here, it is, by far, easier to use the arclength formula you were given)

7. Originally Posted by Jhevon
your diagram is fine. but you do not have a triangle here. what you have is called a sector. notice that one side of the triangle is not straight, the sine and cosine rules do not apply because you don't have a triangle. use topsquark's formula

(if you really want to, you can use the cosine rule, but it is overkill here, it is, by far, easier to use the arclength formula you were given)

So the formula he gave is

S = r(theta)

Since I'm trying to find the angle, it should be

s / r = theta

2450 / 3960 = theta

theta = .618687

Do I use the inverse of Sine to find the angle measurement? Well first of all is that right?

8. Originally Posted by JonathanEyoon
So the formula he gave is

S = r(theta)

Since I'm trying to find the angle, it should be

s / r = theta

2450 / 3960 = theta

theta = .618687

Do I use the inverse of Sine to find the angle measurement? Well first of all is that right?
yes that is right. you only use the inverse sine if there is a sine function you want to undo, there is no sine here, why would you do inverse sine. that is the answer, the angle is in radians in this formula. if you wanted degrees you can convert it, or use a different formula to begin with

9. You know I tried it with the Law of Cosines and ended up doing this

2450^2 = 3960^2 + 3960^2 - 2(3960)(3960)CosC

CosC = (3960^2+3960^2-2450^2) / 2(3960)(3960)

CosC = .808613

CosC^-1 = 36.0394 Degrees

Is this way acceptable as well if I've ignored the section and replaced it with a straight line to create a triangle? Now I'm wondering which is the real answer to the problem . if the angle is in radians how do you make it to degrees?

10. Originally Posted by JonathanEyoon
You know I tried it with the Law of Cosines and ended up doing this

2450^2 = 3960^2 + 3960^2 - 2(3960)(3960)CosC

CosC = (3960^2+3960^2-2450^2) / 2(3960)(3960)

CosC = .808613

CosC^-1 = 36.0394 Degrees

Is this way acceptable as well if I've ignored the section and replaced it with a straight line to create a triangle? Now I'm wondering which is the real answer to the problem
no, because the 2450 is in fact, not a side of the triangle, it is a curved distance (the distance between the cities is measured along the Earth's surface, is it not?). this will only give you an estimate. the actual answer is closer to 35 degrees.

11. thanks alot!!!

12. Originally Posted by JonathanEyoon
thanks alot!!!
you're welcome