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    I need help answering a question:In triangle abc, cosA = 0.5, AB = 6cm, BC = 2x cm, AC = x cm, show that x^2 + 2x - 12 = 0
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    (Original post by Bethanyt5560)
    I need help answering a question:In triangle abc, cosA = 0.5, AB = 6cm, BC = 2x cm, AC = x cm, show that x^2 + 2x - 12 = 0
    What have you tried? Please post your working.

    If you're unsure how to start, try to use the cosine rule and post your working if you get stuck.
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    (Original post by Bethanyt5560)
    I need help answering a question:In triangle abc, cosA = 0.5, AB = 6cm, BC = 2x cm, AC = x cm, show that x^2 + 2x - 12 = 0
    By the cosine rule, a^2 = b^2 + c^2 - 2bc*cos(A).
    Thus (2x)^2 = x^2 + 6^2 - 2*x*6*0.5 -> 4x^2 = x^2 - 6x + 36 -> 3x^2 + 6x - 36 = 0 -> x^2 + 2x - 12 = 0, as required.
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    (Original post by constellarknight)
    By the cosine rule, a^2 = b^2 + c^2 - 2bc*cos(A).
    Thus (2x)^2 = x^2 + 6^2 - 2*x*6*0.5 -> 4x^2 = x^2 - 6x + 36 -> 3x^2 + 6x - 36 = 0 -> x^2 + 2x - 12 = 0, as required.
    I have already told you that full solutions are against the rules. If you have issue with this then please contact a moderator.

    If you want to continue to post full solutions then I politely ask you to post solutions in a different forum.
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    (Original post by constellarknight)
    By the cosine rule, a^2 = b^2 + c^2 - 2bc*cos(A).
    Thus (2x)^2 = x^2 + 6^2 - 2*x*6*0.5 -> 4x^2 = x^2 - 6x + 36 -> 3x^2 + 6x - 36 = 0 -> x^2 + 2x - 12 = 0, as required.
    We aren't allowed to post full solutions. @notnek tried to get her to think about the question, you might have just denied her that.
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    (Original post by constellarknight)
    By the cosine rule, a^2 = b^2 + c^2 - 2bc*cos(A).
    Thus (2x)^2 = x^2 + 6^2 - 2*x*6*0.5 -> 4x^2 = x^2 - 6x + 36 -> 3x^2 + 6x - 36 = 0 -> x^2 + 2x - 12 = 0, as required.
    Thank you so much, what does * mean
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    (Original post by Bethanyt5560)
    Thank you so much, what does * mean
    * = multiply, e.g. 4*3=12. It's just an easier way of writing the multiplication sign on a computer.
 
 
 
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