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January 24th, 2017, 07:06 AM   #1
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Question Problem About Using Cosine Theorem

Find cosine for the lowest and the largest angles in triangle ABC, if AB:BC:AC=5:7:8


I tried to solve using cosine theorem.
But I did not get the answer that is given in the book.
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January 24th, 2017, 07:23 AM   #2
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Will you show your work? What is the book's answer?
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January 24th, 2017, 07:44 AM   #3
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Yeas, of course.
c^2=a^2+b^2-2*a*b*cosC
(5x)^2=(7x)^2+(8x)^2-2*7x*8x*cosC
25x^2=49x^2+64x^2-112x^2cosC
25=113-112cosC
cosC=11/14


The books's answers: 5/6; -0.1
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January 24th, 2017, 08:01 AM   #4
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smallest angle is C ...

$\cos{C} = \dfrac{7^2+8^2-5^2}{2 \cdot 7 \cdot 8} = \dfrac{11}{14}$

largest angle is B ...

$\cos{B} = \dfrac{7^2+5^2-8^2}{2 \cdot 7 \cdot 5} = \dfrac{1}{7}$
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January 24th, 2017, 08:36 AM   #5
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Yeas, these answers are what i got, but in the book they are not the same.
Book's answers are: 5/6; -0.1
I can think that the book's answer is not correct.
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January 24th, 2017, 08:45 AM   #6
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book "answers" are not immune to being incorrect ...
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January 24th, 2017, 09:14 AM   #7
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Thumbs up

Thank you to all helping me to meditate how wonderful is Mathematics, and how Almighty is The God who created all the lows of Mathematics and not only of Mathematics. I am really happy to be a member of this forum. Great.
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January 24th, 2017, 09:33 AM   #8
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It would seem that the problem should give "if AB:BC:AC=5:7:9". Did you misread it?
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January 24th, 2017, 09:49 AM   #9
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No, no. in the book is: 5:7:8
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