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Math Help - Law of Cosines/Ambiguous case questions!

  1. #1
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    Law of Cosines/Ambiguous case questions!

    I'm having a bit of trouble with this. The unit test is tomorrow morning and any help is very much appreciated.

    Please explain how you got your answers if you're willing to help me. I don't think any of these are particularly hard and if I knew how to do these problems, I'd know how to do the rest, so please help me out. >:

    First of all, I don't understand the law of cosines. I know the formula is a^2 = c^2 + b^2 -2cb (CosA), but I don't know how to use or apply it. What are the requirements for LoC (like, I know Law of Sines needs an angle and its corresponding side, etc.).

    What is the Ambiguous Case and how do I use/apply/recognize that?

    And finally, some specific questions I couldn't nail (even though they seem to be rather easy >:

    1) Calculate the area of the quadrilateral shown below, where area = 1/2 (side1) (side1) (SinAngle):

    http://img401.imageshack.us/my.php?image=math1hp5.jpg

    2) Calculate the area of an equilateral triangle with a perimeter of 59.1 cm.

    3) A triangle has sides measuring 11 cm, 14 cm, and 17 cm. Determine the measure of the smallest angle of this triangle (< ---- what does that mean? I thought it just meant what it said - find the measure of the smallest angle, but I got it wrong, so... My answer was: the smallest angle is a, where a is 11 cm and A is 38.2 degrees).

    4) In triangle CDE, the coordinates of C are (4, -3), the coors. of D are (3, 4) and the coors. of E are (0, 0). Determine the measure of angle CED. <-- I have no idea where to even begin, here.

    5) In triangle RST, two angles measure 40 degrees and 80 degrees. One side measures 17 cm. What is the smallest possible side of RST? <--- What does that mean?
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  2. #2
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    First of all, I don't understand the law of cosines. I know the formula is a^2 = c^2 + b^2 -2cb (CosA), but I don't know how to use or apply it.
    You can use the Law of Cosines to find the length of the third side (a) of a triangle when you know the measure of an angle (A) and the lengths of the two sides (b and c) adjacent to that angle.
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  3. #3
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    What are the requirements for LoC (like, I know Law of Sines needs an angle and its corresponding side, etc.).

    What is the Ambiguous Case and how do I use/apply/recognize that?
    The Ambiguous Case occurs when you know two side lengths and one angle measure, and that angle measure's opposite side is the shorter side.
    In this case, there are two possibilities for the measure of the angle opposite the longer side.
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    You can use the Law of Cosines to find the length of the third side (a) of a triangle when you know the measure of an angle (A) and the lengths of the two sides (b and c) adjacent to that angle.
    Oh, I see. So it'd be like:

    In triangle DEF, d = 14.5 cm, e = 8.9 cm, and angle F = 18 degrees. Solve this triangle.

    And your formula would be f^2 = d^2 + e^2 - 2de (CosF)? Or have I gone backwards with it?

    The Ambiguous Case occurs when you know two side lengths and one angle measure, and that angle measure's opposite side is the shorter side.
    So if I had triangle ABC, where A = 30 degrees, c = 15, and a = 14, it'd be amb. case?
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  5. #5
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    Quote Originally Posted by Pipptart View Post
    Oh, I see. So it'd be like:

    In triangle DEF, d = 14.5 cm, e = 8.9 cm, and angle F = 18 degrees. Solve this triangle.

    And your formula would be f^2 = d^2 + e^2 - 2de (CosF)? Or have I gone backwards with it?
    Yes, that formula is correct.

    So if I had triangle ABC, where A = 30 degrees, c = 15, and a = 14, it'd be amb. case?
    Yes, it would be an ambiguous case. Can you determine the possible values for angle C in this case?
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  6. #6
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    I'm not sure. I don't really know what I have to do.

    Is there a formula I have to use?
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  7. #7
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    Quote Originally Posted by Pipptart View Post
    I'm not sure. I don't really know what I have to do.

    Is there a formula I have to use?
    The law of sines can give you the sine of angle C, from which you can determine the possible angles.
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  8. #8
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    Quote Originally Posted by icemanfan View Post
    The law of sines can give you the sine of angle C, from which you can determine the possible angles.
    so SinA/a = SinC/c

    Sin 30/14 = Sin C/15

    SinC (14) = Sin 30 (15)

    SinC = Sin 30 (15)/14

    C = 32 degrees?
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  9. #9
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    Quote Originally Posted by Pipptart View Post
    so SinA/a = SinC/c

    Sin 30/14 = Sin C/15

    SinC (14) = Sin 30 (15)

    SinC = Sin 30 (15)/14

    C = 32 degrees?
    C = 32 degrees or 148 degrees. Remember, in the ambiguous case there are two angles.
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  10. #10
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    Quote Originally Posted by icemanfan View Post
    C = 32 degrees or 148 degrees. Remember, in the ambiguous case there are two angles.
    Why are there two angles? And how did you find 148 degrees?
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  11. #11
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    Quote Originally Posted by Pipptart View Post
    Why are there two angles? And how did you find 148 degrees?
    There are two possible angles because the triangle can be constructed two different ways. To understand this, it might be useful to construct the angle CAP and then set a compass for the length of CB (which is a) and then set the point of the compass at C and draw a circle. The circle will intersect the ray AP at two points.

    148 = 180 - 32.
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  12. #12
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    Quote Originally Posted by icemanfan View Post
    There are two possible angles because the triangle can be constructed two different ways. To understand this, it might be useful to construct the angle CAP and then set a compass for the length of CB (which is a) and then set the point of the compass at C and draw a circle. The circle will intersect the ray AP at two points.

    148 = 180 - 32.
    Ohh, okay. I understand amb. case now.
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