Results 1 to 12 of 12

Math Help - Trigonometry word problem

  1. #1
    Senior Member
    Joined
    Jul 2007
    Posts
    290

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,079
    Thanks
    375
    Awards
    1
    Quote Originally Posted by JonathanEyoon View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by topsquark View Post
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Jhevon View Post
    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.
    Attached Thumbnails Attached Thumbnails Trigonometry word problem-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Jhevon View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    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?
    Attached Thumbnails Attached Thumbnails Trigonometry word problem-untitled.jpg  
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    thanks alot!!!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JonathanEyoon View Post
    thanks alot!!!
    you're welcome
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry word problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 24th 2011, 06:44 PM
  2. Trigonometry word problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 26th 2009, 07:58 PM
  3. Trigonometry Word Problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: January 27th 2009, 05:27 AM
  4. trigonometry word problem
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 7th 2008, 03:14 PM
  5. another trigonometry word problem
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 12th 2008, 11:19 PM

Search Tags


/mathhelpforum @mathhelpforum