Results 1 to 4 of 4

Math Help - bearing problems

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    5

    bearing problems

    I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.

    1.) A plane flew 50km on a bearing 6320', and then flew on a bearing of 15320' for 140km. Find the distance between the starting point and the ending point.

    2.) A boat left the port on a course of 23020' for 1.5hour at 210mph and then at 14020' Edit*-(for 2 hr at 180mph)-. How far is the boat from the port? In what direction must it travel to return to the port?

    3.) City A and City B are located on a north-south line and 34.6 miles apart. An observer on the east of that line finds that the directions of A and B are N2710'W and S6250'W, respectively. How far is he from each city?

    4.) At a certain time, a lighthouse is south of a ship. Thirty minutes later, the ship bears N4020'E from the lighthouse. If the ship is sailing east at 20kph, find the distance of the ship from the lighthouse at each time of observations.

    5.) A ship leaves its home port and sails on a bearing of N3815'E at 24mph. At the same instant. another ship on the same port on a bearing of S5145'E at 28mph. Find the distance between the 2 ships after 4 hrs.

    6.) A surveyor describes a tract of land by saying : starting at a concrete stake, thence N3720'E for 120km, thence S5240'E until due east of the starting point, thence to the starting point. Find the perimeter of the tract.


    Any help would be greatly appreciated.
    Last edited by angeIi28; June 28th 2009 at 12:23 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by angeIi28 View Post
    I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.

    1.) A plane flew 50km on a bearing 6320', and then flew on a bearing of 15320' for 140km. Find the distance between the starting point and the ending point.

    2.) A boat left the port on a course of 23020' for 1.5hour at 210mph and then 14020' at 180mph. How far is the boat from the port? In what direction must it travel to return to the port?

    3.) City A and City B are located on a north-south line and 34.6 miles apart. An observer on the east of that line finds that the directions of A and B are N2710'W and S6250'W, respectively. How far is he from each city?

    4.) At a certain time, a lighthouse is south of a ship. Thirty minutes later, the ship bears N4020'E from the lighthouse. If the ship is sailing east at 20kph, find the distance of the ship from the lighthouse at each time of observations.

    5.) A ship leaves its home port and sails on a bearing of N3815'E at 24mph. At the same instant. another ship on the same port on a bearing of S5145'E at 28mph. Find the distance between the 2 ships after 4 hrs.

    6.) A surveyor describes a tract of land by saying : starting at a concrete stake, thence N3720'E for 120km, thence S5240'E until due east of the starting point, thence to the starting point. Find the perimeter of the tract.


    Any help would be greatly appreciated.
    1. You don't need any Sine or Cosine (except #2!) because all problems are about right triangles. Simple Pythagorean theorem will do.

    2. The problem with your problems is that you have to convert the bearings and courses into a more convenient format of angle measure:
    Courses are given wrt to North N = 0, measured clockwise. Thus
    N = 0
    E = 90
    S = 180
    W = 270

    A course of S6250'W = 180 + 6250' = 24250'
    A course of S5240'E = 180 - 5240' = 17960' - 5240' = 12720'

    3. Draw a rough sketch for each problem.

    4. The problem #2 can't be solved because the traveling time of the 2nd course is missing.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by angeIi28 View Post
    I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.
    ...
    Any help would be greatly appreciated.
    As earboth stated, you don't need to use the trig laws,
    but you may do so as as option.

    For all of the calculations:
    distance x sin(bearing) = y component
    distance x cos(bearing) = x component

    Problem 1):
    50 x sin(6320') = 50 x 0.89363 = 44.68 = y1
    50 x cos(6320') = 50 x 0.4488 = 22.44 = x1

    140 x sin(15320') = 140 x 0.4488 = 62.83 = y2
    140 x cos(15320') = 140 x -0.89363 = -125.11 = x2

     Y_{endingpoint} = y_1 + y_2 = 44.68+62.83 = 107.51

     X_{endingpoint} = x_1 + x_2 = 22.44 + (-125.11) = -102.67

    The distance between the starting point and the ending point:

     \sqrt{ 107.51^2 + (-102.67)^2}

    Since the sin of one bearing equals the cos of the other bearing (except for sign) & the cos of the one bearing equals the sin of the other bearing, you have a right triangle.

     \sqrt{ 50^2 + 140^2}

    Both results should be equal.
    You may need more decimal precision that I have given here.


    NOTE:
    When North is used the sin value will be positive.
    When South is used the sin value will be NEGATIVE.
    That means that the y component will be positive or negative.

    When East is used the cos value will be positive.
    When West is used the cos value will be negative.
    That means that the x component will be positive or negative.

    ---
    NOTE 2:
    You can use the cosine law on this:

    The difference between the two bearings (6320' and 15320') is angle C.

    the distance between the starting & ending:

     c^2 = a^2 + b^2 - 2ab \, \cos C
     c^2 = 50^2 + 140^2 - 2 \times 50 \times 140 \times \cos (153deg20min - 63deg20min)


    Then take the square root of c^2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2009
    Posts
    5

    thanks

    thanks for the help... I still feel uneasy about bearing problems but now I'm beginning to understand it. Thanks again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bearing problems
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: September 11th 2009, 10:16 AM
  2. Bearing Problems
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 14th 2009, 08:55 AM
  3. Bearing word problems help
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 25th 2008, 05:41 AM
  4. Replies: 4
    Last Post: November 10th 2007, 10:22 PM
  5. Trigonometry: bearing problems
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 30th 2007, 08:38 PM

Search Tags


/mathhelpforum @mathhelpforum