Results 1 to 3 of 3

Math Help - The distance between two planes

  1. #1
    Newbie
    Joined
    Jun 2008
    Posts
    23

    The distance between two planes

    The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes?

    Attempted Solution: First, I drew a triangle.

    Then I used the law of cosines.

    c^2 = a^2 + b^2 -2ac(cos \theta)

    Then put in the a, c, and  cos \theta) values.

    c^2 = 17^2 + 20^2 - 2ac(cos 110^o)
    c^2 = 289 + 400 - 680(cos 110^0)
    c^2 = 679.33
    {\sqrt{c^2}} = {\sqrt{679.33}}
    c = 21.06

    So the answer should be 21.06 km. But, according to the answer key, it's 30.36 km. I cannot figure out where I erred. Could someone please help?
    Attached Thumbnails Attached Thumbnails The distance between two planes-untitled.png  
    Last edited by D. Martin; September 20th 2008 at 08:47 PM. Reason: To add diagram
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    The part where you went wrong is that the Law of Cosine you used isn't correct.

    c^2 = 17^2 + 20^2 - 2{\color{red}ac}(cos 110^o)
    It should be ab.

    Let A, B, and C be the vertices of a triangle and a, b, and c be the sides of the same triangle. Side a is opposite vertex A, side b is opposite vertex B, side c is opposite vertex C.

    The Law of Cosines state that:
    a^2 = b^2 + c^2 - 2bc\cos{A}

    b^2 = a^2 + c^2 - 2ac\cos{B}

    c^2 = a^2 + b^2 - 2ab\cos{C}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by D. Martin View Post
    The Problem: One plane is 20 km from the Winnipeg Airport. A larger plane is 17 km from the airport. If the angle between the sightings is 110 degrees, how far apart are the planes?

    Attempted Solution: First, I drew a triangle.

    Then I used the law of cosines.

    c^2 = a^2 + b^2 -2ac(cos \theta)

    Then put in the a, c, and  cos \theta) values.

    c^2 = 17^2 + 20^2 - 2ac(cos 110^o)
    c^2 = 289 + 400 - 680(cos 110^0)
    c^2 = 679.33
    {\sqrt{c^2}} = {\sqrt{679.33}}
    c = 21.06

    So the answer should be 21.06 km. But, according to the answer key, it's 30.36 km. I cannot figure out where I erred. Could someone please help?
    -680cos(110deg) is +232.5737
    It is not -232.5737
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Distance between planes (Proof)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2011, 11:35 AM
  2. Distance between the given parallel planes:
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 28th 2010, 03:30 AM
  3. Shoetest distance between 2 planes
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 11th 2009, 06:22 AM
  4. Distance between two parallel planes
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 13th 2009, 10:18 AM
  5. Distance between two planes
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 13th 2009, 03:42 AM

Search Tags


/mathhelpforum @mathhelpforum