Results 1 to 2 of 2

Math Help - Precalculus angle-side-side, please help!!

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    1

    Precalculus angle-side-side, please help!!

    Hi, I really need help on this problem please. I realize that because it's angle side side, and one is shorter than the other, there will be two sets of solutions because there are two triangles. Please help me and show me what both areas and perimeters are! Thanks!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    If the triangle is shown like that, then there is only one solution.

    If the triangle is not shown like that, if only the two sides and the one angle are mentioned without showing a figure, then there are two solutions.

    Maybe you just showed the figure of one of the two possible solutions.

    For the figure that is shown,

    Let's call the triangle, ABC, where
    >>>angle A is 15 degrees
    >>>AB = 11 ft
    >>>BC = 8 ft

    By Law of Sines,
    8/sin(15deg) = 11/sinA
    Cross multiply,
    8sinA = 11sin(15deg)
    sinA = 11sin(15deg) /8 = 0.355876
    angle A = arcsin(0.355876) = 20.847153 degrees ----**

    angle B = 180 -15 -20.847153 = 144.152847 deg

    By Law of Sines again,
    8/sin(15deg) = AC/sin(144.152847deg)
    AC = 18.1 ft

    So, perimeter of ABC = 11 +8 +18.1 = 37.1 ft -------answer.
    Area = (1/2)(11)(18.1)sin(15deg) = 25.765 sq.ft -----answer.

    --------------------------------
    For the other triangle:

    Again,
    By Law of Sines,
    8/sin(15deg) = 11/sinA
    Cross multiply,
    8sinA = 11sin(15deg)
    sinA = 11sin(15deg) /8 = 0.355876 ---------positive sine value, so A is in the 1st or 2nd quadrant.

    In the 1st quadrant,
    angle A = arcsin(0.355876) = 20.847153 degrees ----used in the above triangle.

    In the 2nd quadrant,
    angle A = 180 -20.847153 = 159.152847 deg -----to be used in this triangle.

    angle B = 180 -15 -159.152847 = 5.847153 deg

    By Law of Sines again,
    8/sin(15deg) = AC/sin(5.847153 deg)
    AC = 3.15 ft

    So, perimeter of ABC = 11 +8 +3.15 = 22.15 ft -------answer.
    Area = (1/2)(11)(3.15)sin(15deg) = 4.484 sq.ft -----answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: November 13th 2011, 02:32 PM
  2. Normal distribution beams laid side by side
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 12th 2011, 03:05 AM
  3. Replies: 0
    Last Post: October 9th 2010, 12:40 PM
  4. Angle and side in a pentagon
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 5th 2009, 05:16 PM
  5. Replies: 1
    Last Post: September 14th 2008, 12:25 PM

Search Tags


/mathhelpforum @mathhelpforum