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Math Help - Law of Sine and Cosine Question

  1. #1
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    Law of Sine and Cosine Question

    C----35----A
    --\---------/
    ---\------/
    ---28--17
    -----\--/
    ------\/
    ------B
    This is what the triangle is suppose to be like.
    Law of sine: (sinA/a)=(sinB/b)=(sinC/c)
    Law of cosines: a2=b2+c2-2bc(cosA)
    b2=a2+c2-2ac(cosB)
    c2=a2+b2-2ab(cosC)
    Capital letters are the angles, while lowercase letters are the side lengths
    a=28, b=35, c=17
    Basically I'm trying to find all the angles of the triangle.

    First I find angle A
    282=352+172-2(35)(17)cosA
    784=1514-1190cosA
    -730=1190cosA
    Angle A is about 52.193 degrees

    Then I find Angle B, but this is where I'm confused at, I try to find it 3 different ways:
    1. Using Law of Sine
    (sin52.193/28)=(sinB/35)
    sinB = 0.988
    sin-1(0.988)=81.115 degrees=Angle B

    2. Using Law of Sine
    (sin52.193/28)=(sinC/17)
    sinC = 0.479
    sin-1(0.479)=28.620
    180-52.193-28.620=99.187=Angle B

    3. Using Law of Cosine
    352=282+172-2(28)(27)(cosB)
    1225=1073-2(952cosB)
    152=1904cosB
    0.0799=cosB
    cos-1(0.0799)=85.4162=Angle B

    So basically my teacher got somewhere around 99 degrees for angle B, which is close to what I got for solution #2
    But I was wondering, how do you find out which one to use, because they all seem to work
    Thanks!
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  2. #2
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    Re: Law of Sine and Cosine Question

    Quote Originally Posted by Chaim View Post
    C----35----A
    --\---------/
    ---\------/
    ---28--17
    -----\--/
    ------\/
    ------B
    This is what the triangle is suppose to be like.
    Law of sine: (sinA/a)=(sinB/b)=(sinC/c)
    Law of cosines: a2=b2+c2-2bc(cosA)
    b2=a2+c2-2ac(cosB)
    c2=a2+b2-2ab(cosC)
    Capital letters are the angles, while lowercase letters are the side lengths
    a=28, b=35, c=17
    Basically I'm trying to find all the angles of the triangle.

    First I find angle A
    282=352+172-2(35)(17)cosA
    784=1514-1190cosA
    -730=1190cosA
    Angle A is about 52.193 degrees

    Then I find Angle B, but this is where I'm confused at, I try to find it 3 different ways:
    1. Using Law of Sine
    (sin52.193/28)=(sinB/35)
    sinB = 0.988
    sin-1(0.988)=81.115 degrees=Angle B

    2. Using Law of Sine
    (sin52.193/28)=(sinC/17)
    sinC = 0.479
    sin-1(0.479)=28.620
    180-52.193-28.620=99.187=Angle B

    3. Using Law of Cosine
    352=282+172-2(28)(27)(cosB)
    1225=1073-2(952cosB) mistake in this line ... 2(28)(27) = 952 ... you multiplied by 2 once too many times
    152=1904cosB
    0.0799=cosB
    cos-1(0.0799)=85.4162=Angle B

    So basically my teacher got somewhere around 99 degrees for angle B, which is close to what I got for solution #2
    But I was wondering, how do you find out which one to use, because they all seem to work
    as far as your two solutions for angle B, note that if b^2 > a^2 + c^2 , then angle B is obtuse
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  3. #3
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    Sheffield England
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    Re: Law of Sine and Cosine Question

    Also cosB is negative ( -152/952) so B is obtuse as you say (B=99.2)
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