Results 1 to 5 of 5

Math Help - [SOLVED] cosine rule help!

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    5

    Angry [SOLVED] cosine rule help!

    I have racked my brain trying to think of what I've done wrong, but I just can't.

    Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

    x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
    x^2=91.53-90.72Xcos17
    x^2=91.53-86.76
    x^2=4.77
    x=2.18


    Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

    xxx
    marika
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,803
    Thanks
    114
    Quote Originally Posted by xch3m_is_l0v3x View Post
    I have racked my brain trying to think of what I've done wrong, but I just can't.

    Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

    x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
    x^2=91.53-90.72Xcos17
    x^2=91.53-86.76
    x^2=4.77
    x=2.18


    Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

    xxx
    marika
    I really don't understand where from you've got these values

    1. Distance walked by Jim: 8\ s \cdot 1.4\ \frac ms = 11.2\ m

    2. Distance walked by Tony: 8\ s \cdot 1.7\ \frac ms = 13.6\ m

    3. Angle included by the two directions: 50

    4. Therefore the distance between them will be:

    x^2=11.2^2+13.6^2-2\cdot 11.2 \cdot 13.6 \cdot \cos(50^\circ)

    I've got x\approx 10.7\ m
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Cosine Rule

    Hello xch3m_is_l0v3x
    Quote Originally Posted by xch3m_is_l0v3x View Post
    I have racked my brain trying to think of what I've done wrong, but I just can't.

    Jim and Tony leave the same point at the same time with Jim walking away at the speed of 1.4m/s and tony at a speed of 1.7m/s, the two directions of travel making a angle of 50 degrees with eachother. If they continue on the same straight line paths how far apart are they after 8 seconds?

    x^2=6.3^2+7.2^2-2(6.3)(7.2)cos17
    x^2=91.53-90.72Xcos17
    x^2=91.53-86.76
    x^2=4.77
    x=2.18


    Can someone please help me? I want to know what I've done wrong! and thank you for taking the time to read this post =]

    xxx
    marika
    Thanks for showing us your working, but I don't understand where you get your numbers from. Is this the same question? After 8 sec Jim has walked 1.4 x 8 = 11.2 m and Tony 1.7 x 8 = 13.6 m. The angle between their paths is 50 degrees. So

    x^2 = 11.2^2 + 13.6^2 - 2\times 11.2 \times 13.6 \cos 50^o

    Grandad

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    Posts
    2
    hi i like to use the sine law
    thats sine equals opposite over hypothenuse
    thats sine 50 times 13.6 (8x1.7)
    sine 50 = 0.766
    0.766x13.6=10.42

    hope you agree
    thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Cosine Rule

    Hello nardu
    Quote Originally Posted by nardu View Post
    hi i like to use the sine law
    thats sine equals opposite over hypothenuse
    thats sine 50 times 13.6 (8x1.7)
    sine 50 = 0.766
    0.766x13.6=10.42

    hope you agree
    thanks
    You can only use sine = \frac{\text{opposite}}{\text{hypotenuse}} when you have a right-angled triangle, and I'm afraid that the triangle in this question isn't right-angled. In a case like this, then, the formulae you use are rather different. You'll find a summary of them here: Sine and Cosine Formulae

    Good luck.

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Sine and Cosine Rule
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 6th 2010, 02:51 AM
  2. Help with cosine rule!
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 17th 2009, 02:19 AM
  3. cosine rule
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 15th 2009, 02:20 AM
  4. [SOLVED] Cosine rule 2
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 10th 2009, 04:38 AM
  5. Cosine Rule
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: February 4th 2009, 10:37 AM

Search Tags


/mathhelpforum @mathhelpforum