Please help me with a math (trig) assignment: finding angle

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  1. #1
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    Please help me with a math (trig) assignment: finding angle

    I have to find the measure of the largest angle in this triagnle. The triangle has sides with a measure of 121m, 173m, and 194m. It is not a right triangle because 121^2 + 173^2 != 194^2. I am not given any angles. I know the sine law, but I am not sure what steps I should take from here on out. If anyone could 'ask me the right questions' to show me how to get back on track I'd highly appreciate it.

    Here's a picture of what I have so far:

    EDIT:
    sinC = h / 121
    sinA = h / 121
    194sinA = 121sinC (these both solve for h)
    x^2 + h ^2 = 121^2
    (173 - x) ^2 + h ^2 = 194 ^2
    I really can't think of anything else. Is this even possible?
    Attached Images Attached Images  
    Last edited by Silvercord; 03-25-2003 at 06:07 PM.

  2. #2
    Registered User Commander's Avatar
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    if you use cosine instead of sine, and the cosine law, it might work out.

    EDIT: i got
    A .= 38
    B .= 93 <==
    C .= 49
    Last edited by Commander; 03-25-2003 at 07:55 PM.
    oh i'm sorry! i didn;t realize my fist was rushing to meet ur face!

    MSN :: commander_android@hotmail.com []*[]

  3. #3
    i want wookie cookies the Wookie's Avatar
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    use the law of cosines

    cos C = (a^2 + b^2 - c^2)/2ab

    that will give u the measure of angle C, then u can use the law of sine to find the other angles:

    sin A/a = sin B/b = sin C/c
    Last edited by the Wookie; 03-25-2003 at 07:55 PM.

  4. #4
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    yea it should work out for you...cosine law should word fine...

    remember that c^2 = h^2 +(173-x)^2

    and that c^2 = a^2 + b^2 -28(a)(173)cosC...I think...

    ...let us know if you have any problems...I might not be able to code much but math I can do...

    ...but I think the guys have you covered here...
    ..."GSXR1300"...

  5. #5
    5|-|1+|-|34|) ober's Avatar
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    In the future, you can try here: Clicky

    There's quite a few math people there and they're willing to help.

  6. #6
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    Ok thanks you guys. I haven't been taught the law of cosines yet so I dont' know why they've given us this problem, but i'll look into solving it

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    Hi commander, I actually got a different answer, and I'm not sure who is right. I used both of these following equations ( the first is how I would've done it on my own and the second is what was shown to me in this thread, but they're the same)

    cosC = (c^2 - (a^2 + b^2)) / -2(ab)
    cosC = (c^2 + b^2 - c^2) / 2ab

    here is what I got for angles:
    angle C = 80.46
    angle B = 61.58
    angle A = 37.96

    EDIT: I'm right I think, look at this site, move every one of my angles in the I posted picture above clockwise to get the same placement they did

    type in 194 for a, 121 for b, and 173 for c, then click angle C and you should get 61.57

    http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html
    Last edited by Silvercord; 03-26-2003 at 04:19 PM.

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