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Math Help - Urgent Trig Help law of sines

  1. #1
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    Urgent Trig Help law of sines

    Hi!

    I am learning the Law of Sines, and have no idea what to do. I have notes from my class, but the whole multiple solutions thing is throwing me. I do, however, understand the basics about the law of Sines.

    Here is a problem I don't understand- it is an Ambiguous case:

    Determine the number of possible solutions. If a solution exists, solve the triangle.
    Triangle ABC
    C=17 degrees a= 10 units c= 11 units

    How would I go about finding <A and side B so that I can check to see how many solutions?

    Thanks,

    Lyrixa
    Last edited by Lyrixa; November 25th 2006 at 03:44 PM.
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  2. #2
    Member Ranger SVO's Avatar
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    Lets pretend for a minute that we have a triangle with three sides measuring 5-inches, 8-inches and 2-inches.

    No triangle exists for these measurements. Why?

    Because side A + side B must be greater than side C ( I always pick the longest side for C
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  3. #3
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    Quote Originally Posted by Lyrixa View Post
    Hi!

    I am learning the Law of Sines, and have no idea what to do. I have notes from my class, but the whole multiple solutions thing is throwing me. I do, however, understand the basics about the law of Sines.

    Here is a problem I don't understand- it is an Ambiguous case:

    Determine the number of possible solutions. If a solution exists, solve the triangle.
    Triangle ABC
    C=17 degrees a= 10 units c= 11 units

    How would I go about finding <A and side B so that I can check to see how many solutions?

    Thanks,

    Lyrixa
    We know by the law of sines,
    \frac{11}{\sin 17^o}=\frac{10}{\sin A}
    Cross multiply,
    11\sin A=10\sin 17^o\approx 2.92
    Divide both sides by 11,
    \sin A=.265
    Using the inverse sine function on calculator (that is \boxed{\sin^{-1}}).
    You get,
    A\approx 15^o
    But wait!
    There is still another possible angle (in Second Quadrant, because sine is positive in first and second).
    So,
    A=180^o -15^o=165^o
    So there are 2 possible triangles.
    And those are the angles that angle A can have.
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  4. #4
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    Quote Originally Posted by Ranger SVO View Post
    Lets pretend for a minute that we have a triangle with three sides measuring 5-inches, 8-inches and 2-inches.

    No triangle exists for these measurements. Why?

    Because side A + side B must be greater than side C ( I always pick the longest side for C
    Okay... but if you were only given two sides, how would you find the value of the third so that you can see if sideA+sideB would be greater than sideC? Like if you were only given 5 inches and 2 inches?
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    Quote Originally Posted by ThePerfectHacker View Post
    We know by the law of sines,
    \frac{11}{\sin 17^o}=\frac{10}{\sin A}
    Cross multiply,
    11\sin A=10\sin 17^o\approx 2.92
    Divide both sides by 11,
    \sin A=.265
    Using the inverse sine function on calculator (that is \boxed{\sin^{-1}}).
    You get,
    A\approx 15^o
    But wait!
    There is still another possible angle (in Second Quadrant, because sine is positive in first and second).
    So,
    A=180^o -15^o=165^o
    So there are 2 possible triangles.
    And those are the angles that angle A can have.
    Okay, figuring the value of A1 now makes sense, but the only way you know that there is another triangle is because sine is positive in both the 1st and 2nd Quadrants? Wouldn't that make every triangle have 2 solutions?
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  6. #6
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    Quote Originally Posted by Lyrixa View Post
    Okay, figuring the value of A1 now makes sense, but the only way you know that there is another triangle is because sine is positive in both the 1st and 2nd Quadrants? Wouldn't that make every triangle have 2 solutions?
    No!

    Say you end up with the following, after some algebra,
    \sin A=2
    There is no solution!
    Because the largest value of sine is 1.
    So there is no triangle.

    WAIT!!!
    UPDATE

    I made a mistake so the value of A is:
    15,165
    But it cannot be 165
    Because the other angle you are given is,
    17
    So when you add them there is no way you can get 180 degree like in a triangle.
    So it can only have 1 angle.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    No!

    Say you end up with the following, after some algebra,
    \sin A=2
    There is no solution!
    Because the largest value of sine is 1.
    So there is no triangle.

    WAIT!!!
    UPDATE

    I made a mistake so the value of A is:
    15,165
    But it cannot be 165
    Because the other angle you are given is,
    17
    So when you add them there is no way you can get 180 degree like in a triangle.
    So it can only have 1 angle.
    Okay... That makes a lot more sense... so on the second triangle, 180 -angleA2 -angleB = angleC, and there is only a second triangle if angleC is positive?
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  8. #8
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    Quote Originally Posted by Lyrixa View Post
    Okay... That makes a lot more sense... so on the second triangle, 180 -angleA2 -angleB = angleC, and there is only a second triangle if angleC is positive?
    That is how you use it to find the third angle.
    But if you get the angle as a negative answer like here it means we have a contradiction, that there is no such triangle with such a side.
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  9. #9
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    Okay... I think I understand all of this now. Thank you so much for your help!!!!!

    Lyrixa
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  10. #10
    Member Ranger SVO's Avatar
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    Only one triangle is possible.

    I hope someone can give you a better explination
    Sin(C)/c = Sin(A)/a so we have Sin(17)/11 = Sin(A)/10

    We should get (10*Sin(17))/11 = Sin(A)
    Sin(A) should equal .266 which will get us an angle of about 15.414-degrees

    Now look at the picture, Line A and Line C are fixed at 10 and 11. It is not possible for me to move line C and make another triangle.
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  11. #11
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    Quote Originally Posted by Ranger SVO View Post
    Only one triangle is possible.

    I hope someone can give you a better explination
    Sin(C)/c = Sin(A)/a so we have Sin(17)/11 = Sin(A)/10

    We should get (10*Sin(17))/11 = Sin(A)
    Sin(A) should equal .266 which will get us an angle of about 15.414-degrees

    Now look at the picture, Line A and Line C are fixed at 10 and 11. It is not possible for me to move line C and make another triangle.
    I see that... You can't move line C because it is fixed at 11 and would be too short!! Thanks for using the picture, I guess I lean more toward being a visual learner.

    ~Lyrixa
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  12. #12
    Member Ranger SVO's Avatar
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    I cannot stop drawing pictures. In the sketch below I think you can see why two triangles are possible. Angle C = 26.6-degrees with Side C = 4.5-inches and side A is 9-inches. I can move side C and maintain its fixed length.

    Angle A can equal 63.2-degrees or 180-62.8 = 117.2-degrees
    Attached Thumbnails Attached Thumbnails Urgent Trig Help law of sines-sines.jpg  
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  13. #13
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    Smile

    Quote Originally Posted by Ranger SVO View Post
    I cannot stop drawing pictures. In the sketch below I think you can see why two triangles are possible. Angle C = 26.6-degrees with Side C = 4.5-inches and side A is 9-inches. I can move side C and maintain its fixed length.

    Angle A can equal 63.2-degrees or 180-62.8 = 117.2-degrees
    Thank you sooo much!! Because of you and ThePerfectHacker, I was able to finish the problems on the Law of Cosines with ease!

    Thanks Again!!

    Lyrixa

    P.S. The pictures are awesome, and a big help!
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  14. #14
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    Hello, Lyrixa!

    Some of your statements are contradictory . . .


    I am learning the Law of Sines, and have no idea what to do.
    I have notes from my class, but the whole multiple solutions thing is throwing me.
    Don't blame the multiple solutions . . . You can still solve the triangle, right?

    I do, however, understand the basics about the Law of Sines. You do?

    Determine the number of possible solutions. If a solution exists, solve the triangle.

    Triangle ABC: . C = 17^o,\;a = 10,\;c= 11

    How would I go about finding <A and side B?
    But you said you understand the basics of the Law of Sines.
    Why can't you find angle A and side B?

    We have: . \boxed{\begin{array}{ccc}A\:= \\B\:=\\C\:=\end{array}<br />
\begin{array}{ccc}\boxed{\quad} \\ \boxed{\quad}\\ 17^o\end{array}<br />
\begin{array}{ccc}a\:=\\b\:=\\c\:=\end{array}<br />
\begin{array}{ccc}10 \\ \boxed{\quad}\\11\end{array}}

    We want angle A.

    Law of Sines: . \frac{\sin A}{a} \:= \:\frac{\sin C}{c}\quad\Rightarrow\quad \sin A \:= \:\frac{a\sin C}{c}

    So we have: . \sin A \:= \:\frac{10\sin17^o}{11} \:\approx\:0.2658

    Then: . A \:=\:\sin^{-1}(0.2658) \:\approx\:15.4^o . . . or 164.6^o **


    If A = 15.4^o, then  B \,=\,180^o - 15.4^o - 17^o \:=\:147.6^o

    And we have: . \boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}<br />
\begin{array}{ccc}15.4^o \\ 147.6^o \\ 17^o\end{array}<br />
\begin{array}{ccc} a\:=\\b\:=\\c\:=\end{array}<br />
\begin{array}{ccc}10 \\ \boxed{\quad}\\11\end{array}}

    Then: . \frac{b}{\sin B} \:=\:\frac{c}{\sin C}\quad\Rightarrow\quad b \:=\:\frac{c\sin B}{\sin C}

    So we have: . b\:=\:\frac{11\sin147.6^o}{\sin17^o} \:\approx\:20.2

    And we have solved the triangle: . \boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}<br />
\begin{array}{ccc}15.4^o \\ 147.6^o \\ 17^o\end{array}<br />
\begin{array}{ccc} a\:=\\b\:=\\c\:=\end{array}<br />
\begin{array}{ccc}10 \\ 20.2\\11\end{array}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Every time you use the \boxed{\sin^{-1}} key, there are two possible answers.

    One is the answer on your calculator screen, always an acute angle.
    . . The other is its supplement (subtract from 180)

    We must check to see if the other angle is a possible answer.
    . . In this case, no.

    If A = 164.6^o, our chart would looks like this:

    . . \boxed{\begin{array}{ccc}A\:=\ \\ B\;= \\ C\:=\end{array}<br />
\begin{array}{ccc}164.6^o \\ \boxed{\quad}\\ 17^o\end{array}<br />
\begin{array}{ccc} a\:=\\b\:=\\c\:=\end{array}<br />
\begin{array}{ccc}10 \\ \boxed{\quad}\\11\end{array}}

    We see that A + C \:=\:181.6^o ... which is impossible.

    Therefore, this triangle is not ambiguous; there is only one solution.

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