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# Need help!! Watch

1. 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
2. (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

If you're unsure how to start, try to use the cosine rule and post your working if you get stuck.
3. (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.
4. (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.
5. (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.
6. (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
7. (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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