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How to solve triangle: all sides given - arccosine doesn't work! Watch

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    Hello (again)

    I have a triangle, with all sides given: 10, 20 and 25.
    So I use the Cosine Rule to fine an angle.

    So I solve: 25^2 = 10^2 + 20^2 - 2 (10)(20) × Cos (A)
    I end up getting: 6.25 = Cos (A)
    But I can't use ArcCosine, because 6.25 is greater than 1(?).

    How does the book expect me to solve this??
    Until now it tells me that I should use the Cosine Rule.

    Many thanks (again) for any help
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    (Original post by the81kid)
    Hello (again)

    I have a triangle, with all sides given: 10, 20 and 25.
    So I use the Cosine Rule to fine an angle.

    So I solve: 25^2 = 10^2 + 20^2 - 2 (10)(20) × Cos (A)
    I end up getting: 6.25 = Cos (A)
    But I can't use ArcCosine, because 6.25 is greater than 1(?).

    How does the book expect me to solve this??
    Until now it tells me that I should use the Cosine Rule.

    Many thanks (again) for any help
    You may want to work through the rearrangement again, you seem to have gone wrong somewhere.
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    (Original post by the81kid)
    Hello (again)

    I have a triangle, with all sides given: 10, 20 and 25.
    So I use the Cosine Rule to fine an angle.

    So I solve: 25^2 = 10^2 + 20^2 - 2 (10)(20) × Cos (A)
    I end up getting: 6.25 = Cos (A)
    But I can't use ArcCosine, because 6.25 is greater than 1(?).

    How does the book expect me to solve this??
    Until now it tells me that I should use the Cosine Rule.

    Many thanks (again) for any help
    You shouldn't be getting 6.25. Have another go and post your working in full if you get the same thing.
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    You should get cosA = -0.3125 I think. Check your mental maths/what you typed into your calculator- the cosine rule always works! x
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    You should get something like \frac{10^2+20^2-25^2}{2*10*20} which gives a different answer to what you obtained.

    cos(A) = \frac{b^2+c^2-a^2}{2bc} is what you should use
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    (Original post by khaleesi98)
    You should get cosA = -0.3125 I think. Check your mental maths/what you typed into your calculator- the cosine rule always works! x
    Ah, I wrote too soon, I think I've found the stupid mistake I've made...
    Sorry guys! : (
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    (Original post by the81kid)
    Ah, I wrote too soon, I think I've found the stupid mistake I've made...
    Sorry guys! : (
    Haha don't worry it happens to everyone 😊
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    (Original post by The-Spartan)
    You should get something like \frac{10^2+20^2-25^2}{2*10*20} which gives a different answer to what you obtained.

    cos(A) = \frac{b^2+c^2-a^2}{2bc} is what you should use
    Yes, now it's working. I'm very sorry. I just came off a one hour headache with the previous problem (which turned out to be wrong because of the Ambiguous Case of Law of Sines which I'd never heard of before). : / My head's exhausted and I just put this problem in wrong.

    Sorry again, and thanks to all for the replies.
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    (Original post by the81kid)
    Yes, now it's working. I'm very sorry. I just came off a one hour headache with the previous problem (which turned out to be wrong because of the Ambiguous Case of Law of Sines which I'd never heard of before). : / My head's exhausted and I just put this problem in wrong.

    Sorry again, and thanks to all for the replies.
    No need to be sorry!! The ambiguous law of sines caught me out once too :P
    In those ambiguous cases, just use that the sum of angles is 180 for future reference :P
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    (Original post by The-Spartan)
    No need to be sorry!! The ambiguous law of sines caught me out once too :P
    In those ambiguous cases, just use that the sum of angles is 180 for future reference :P
    I'll definitely remember it after this afternoon!
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    (Original post by the81kid)
    I'll definitely remember it after this afternoon!
    That is the ideal case of practice makes perfect :P
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    Perimeter = sum sides
    Use Heron formula to find area and heights
    Use asin. example: A=asin(hc/b)
    Draw: triancal.esy.es/?lang=en&tip=2&a=10&b=20&c=25
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