Difference between law of sine and cosine

Discussion in 'Math' started by sairfan1, Apr 17, 2013.

  1. sairfan1

    Thread Starter Member

    May 24, 2012
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    hi,

    I'm learning trigonometry, i want to know difference between law of sine and law of cosine, while learning it is mentioned that both are used for all types of triangles, but i get different answers for same triangle calculated by both laws, plz help.
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    They are the same sequence of numbers, but go in opposite directions for a 0º to 90º value input.

    Try this.
    Take your calculator and put in the number 30 and get the sin value.
    Now take the number 60 and get the cosine value.

    Both are .5

    one is 30 degrees away from 0º(sin), and the other is 30 degrees away from 90º(cos)

    This works for any set of angles. if you have the sin of 25º, it will be equal to the cosine of 65º(90-25)

    Same for the sin of 10º, which will be equal to the cosine of 80º(90-10)


    In a right angle triangle the sin of an angle will be the length of the OPPOSITE side divided by the length of the HYPOTENUSE.

    In a right triangle the cosine of an angle will be equal to the length of the ADJACENT side divided by the length of the HYPOTENUSE
     
  3. amilton542

    Active Member

    Nov 13, 2010
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    It sounds like the ambiguous case, but only the sine law will produce two non-congruent triangles given two sides and not the angle enclosed because  sin(x) \equiv sin(180 - x)
     
    Last edited: Apr 17, 2013
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Perhaps you would like to expand on this?
     
  5. amilton542

    Active Member

    Nov 13, 2010
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    Well, I'm just going with my gut instinct and assuming it's an oblique triangle with two given sides and not the angle enclosed. So (working with a few simple numbers): -

     \frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}

     \frac{3}{sin(A)}=\frac{2}{sin(10^{o})}

     \Rightarrow \angle A = arcsin\frac{3sin(10^{o})}{2}

     \Rightarrow \angle A = 180^{o} - arcsin\frac{3sin(10^{o})}{2} \Rightarrow\ Quad\ II

    Now,

     \angle C = 170^{o} - \left(arcsin\frac{3sin(10^{o})}{2}\right)

    Or,

     \angle C = arcsin\frac{3sin(10^{o})}{2} - 10^{o}

    In such a case, both results appear logical.

    Now,

     \frac{c}{sin\left(170^{o} - \left(arcsin\frac{3sin(10^{o})}{2}\right)\right)} = \frac{2}{sin(10^{o}}

     \Rightarrow c \approx 4.885m

    Or,

     \frac{c}{sin\left(\left(arcsin\frac{3sin(10^{o})}{2}\right) - 10^{o}\right)} = \frac{2}{sin(10^{o}}

     \Rightarrow c \approx 1.023m

    So this gives two non-congruent triangles.

    1st case)

    a = 3m,  \angle (A) \approx 15.098^{o}
    b = 2m,  \angle (B) = 10^{o}
    c = 4.885m  \angle (C) \approx 154.902^{o}

    2nd case)

    a = 3m,  \angle (A) \approx 164.901^{o}
    b = 2m,  \angle (B) = 10^{o}
    c = 1.023m,  \angle (C) \approx 5.098^{o}

    Observe, the largest and shortest side are directly opposite the largest and smallest angles in either case.

    Welcome to the ambiguous case.

    Recall, if all three angles are given then the cosine rule can be given.

    However, because the OP said he's having difficulty with both, I just assumed it was the ambiguous case.
     
  6. amilton542

    Active Member

    Nov 13, 2010
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    Woops. I've made a few Latex errors.
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    OK here is a sketch to help.

    A triangle has six variable, three sides and three angles. Three of these variables, including at least one side, must be known to solve the triangle.

    As to the difference between the sin rule and the cos rule.

    Both connect four variables. Given three you can solve for the fourth.
    Further application yields the fifth and sixth variables.

    The sine rule connects two angles and two sides.

    The cosine rule connects three sides and one angle.

    Using the standard notation a, b, c for the sides and A, B, C for the angles as in my fig 1 in the sketch we can see that in fig2 that if we are given two sides a and c and the angle A, which is not included between them, a difficulty can arise because there may be two possible triangles that can be drawn to satisfy these three values.

    If we draw line AC = b so we position A and C and then draw a line from A at given angle A towards B we have a potential difficulty. We do not yet know where B is since we do not know the length of side A (=c) or angle C.

    If we take a compass and draw a circle centred at C radius = side BC (=a) this will cut line AB produced as shown in two places.

    This gives two possible solutions that may be found by using the cosine rule and solving the quadratic equation that the it leads to, since quadratic equations have two solutions.

    Incidentally

    It is impossible to solve any triangle by any rule unless at least one side is given.

    Does this help?


    Edit : please note I have corrected the mistake in identifying the unknown variable in the attachment. Sorry folks.
     
    Last edited: Apr 18, 2013
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  8. WBahn

    Moderator

    Mar 31, 2012
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    Did your question get answered?

    If not, please post an example of where you are getting different answers so that we can look things over and see what, if anything, you are doing wrong. It may just be a math blunder or it could be roundoff error.
     
  9. sairfan1

    Thread Starter Member

    May 24, 2012
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    thank a lot to all of you, because of time difference my response was late, thanking you again.
     
  10. amilton542

    Active Member

    Nov 13, 2010
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    Oops! I was supposed to say all three sides , not angles.


    Given two sides and one of the angles, you can now solve for the third side.

    Should this not be,"If we are given two sides a and b and the angle A?" Or I have misinterpreted your problem.

    If I havn't, it's very impressive. A much nicer approach as opposed towards rigorous computation for a proof.



    Absolutely. I love geometric proof.
     
  11. WBahn

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    Mar 31, 2012
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    They are equivalent. Basically, you are given one side, the angle opposite it, and one of the other two sides (doesn't matter which).
     
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  12. studiot

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    That is what I said in the attachment. Unfortunately I originally referenced the wrong variable in the last line but that is now corrected.

    Note that if you have only one known side you cannot use the cos rule, you must use the sin rule (or get into solving a pair of simultaneous quadratic equations).
     
  13. amilton542

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    Nov 13, 2010
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    Have you got anymore triangle problems you would like to share?
     
  14. studiot

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    If you mean me, how's your resection?
     
  15. amilton542

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    Nov 13, 2010
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    Resection? What does that mean?

    I've got: -

    "Survey. a technique of ascertaining the location of a point by taking bearings from the point on two other points of known location. "

    Try me. I'll give it my best shot.
     
  16. studiot

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    Yes resection involved fixing one's position by measuring angles between sightings on three (or more) known points.

    The resection I was thinking of is associated with the name Tienstra and involves solving the triangles by the cotangent rule.

    Bearing resection is used principally by navigators and substitutes the north pole for one of the points by using bearings, not angles.

    Other interesting triangle questions include Napoleon's Theorem (yes the 1812 man) and Malfatti's problem.
     
  17. WBahn

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    Mar 31, 2012
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    But note that using the sin rule exposes you to potential fines and penalties, except on Capitol Hill or in certain western states (but it exposes you to other nasty things even there). :D:p
     
  18. studiot

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    That's why I hideout in The Nations, where the penalty is having to drink an extra quart of moonshine. :D
     
  19. amilton542

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    Nov 13, 2010
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    What's your definition of a bearing?

    I always interpret a bearing from North, clockwise.

    Let's say I'm given a bearing of  \left(\frac{\pi}{6}\right)^{c}

    Then I would depict this as an angle of  \left(\frac{\pi}{3}\right)^{c} from the terminal side of Quad I.
     
  20. studiot

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    I agree with all this but didn't catch your point.
     
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