Results 1 to 4 of 4

Math Help - Angles of Elevation?

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    74

    Angles of Elevation?

    Ok. I missed class today so didnt quite get the lesson via math tutorial.

    The question is:

    When a 757 passenger jet begins its descent to the Ronald Reagan International Airport in Dc, it is 3900 ft from the ground. its angle od descent is 6 degrees.

    a. what is the plane's ground distance to the airport?


    b. how far must the plane fly to reach the runway?

    idk how to do it from the tutorial this is what i did....

    cos 6 = 3900
    6 = 3921.5

    is that how im suppose to folllow thru and do?
    please any suggestions
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    982
    part (a)

    \tan(6^{\circ}) = \frac{opposite}{adjacent} = \frac{3900}{x}

    solve for x, the horizontal distance.

    part (b)

    \sin(6^{\circ}) = \frac{opposite}{hypotenuse} = \frac{3900}{z}

    solve for z, the distance the airplane must fly to reach the runway.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,908
    Thanks
    766
    Hello, topaz192!


    When a passenger jet begins its descent to the airport,
    it is 3900 ft from the ground. Its angle of descent is 6.
    First, make a good sketch.
    Code:
        A * - - - - - - - - - - - H
          :   *  6
          :       *
     3900 :           *
          :               *
          :               6  *
        B * - - - - - - - - - - - * C

    The jet is at A. . AB = 3900 feet. .The airport is at C.
    \angle HAC = \angle ACB = 6^o




    a) What is the plane's ground distance to the airport?
    We want distance BC.

    In right triangle ABC \!:\;\;\tan6^o = \frac{3900}{BC} \quad\Rightarrow\quad BC \:=\:\frac{3900}{\tan6^o} \:=\:37,106.02137

    Therefore: . BC \;\approx\;37,106.0\text{ feet}




    b) How far must the plane fly to reach the runway?
    We want distance AC.

    In right triangle ABC\!:\;\;\sin6^o \:=\:\frac{3900}{AC} \quad\Rightarrow\quad AC\:=\:\frac{3900}{\sin6^o} \;=\;37,310.41171

    Therefore: . AC \;\approx\;37,310.4\text{ feet}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    Posts
    74
    man i have the worst problem trying to member the differences of cos, tan, and sin.

    i did it all wrong cuz i did cos.

    thanks yall!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Angles of Elevation and Depression
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 2nd 2010, 02:28 PM
  2. Angles of Elevation and Depression Help!
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 1st 2010, 09:42 PM
  3. Angles of Elevation and Deppresion
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 20th 2010, 05:57 AM
  4. Angles of Elevation and Depression??
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 11th 2009, 02:14 PM
  5. angles of elevation and depression
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 28th 2007, 09:37 PM

Search Tags


/mathhelpforum @mathhelpforum