trigonometry, law of cosines, algebra

Discussion in 'Math' started by ninjaman, Aug 4, 2016.

May 18, 2013
306
1
Hello,
I am having a little problem with my trigonometry and law of cosines. The book I have explains the trigonometry but I think it is the algebra that I am having a problem with.
I have,

I would have values for "C", "B", "cosc" but not "A"
I get stuck when I don't have "A", I work out an answer where I have,

So I end up getting "2Bcosc" but not "A", the same on the other side. I think it is just something to do with rearranging.
Any help would be great thanks

2. Kermit2 AAC Fanatic!

Feb 5, 2010
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965
2ABcosC - B(2) + C(2) = A(2)

3. Alec_t AAC Fanatic!

Sep 17, 2013
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Hint: You have all the values you need for solving a quadratic equation of the form ax^2 + bx +c = 0.

4. Kermit2 AAC Fanatic!

Feb 5, 2010
3,850
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But his "C" term is also squared.

May 18, 2013
306
1
there are some questions in the book that involve the quadratic equation which i have some problems with at the moment as i am focusing on trigonometry.
thanks for the replies, all the beat
simon

6. wayneh Expert

Sep 9, 2010
12,382
3,241
I don't quite understand your question. Are you saying you want to solve explicitly for A, given the others?

7. Papabravo Expert

Feb 24, 2006
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Let A = x, then,

$x^{2} \; -\; 2 \cdot B \cdot cos(c)\cdot x\; +\; (B^{2}\; -\; C^{2}) = 0$

substitute for B, C, cos(c) and solve for A. Negative values for A should be ignored.

Alec_t likes this.

May 18, 2013
306
1
how do i pull out X^2 and X?
this is the problem. i dont know how to solve for x.
i have
C = 13
B = 8
cos c 120
i have to find A
i am using C^2 = A^2-2ABcosc+B^2
i end up with -8A = A^2-105
then -105 over -8, or square A^2-105 to get A route7 route15 after that i get stuck.
any help is appreciated.
thanks

9. wayneh Expert

Sep 9, 2010
12,382
3,241
Remind yourself how to solve a standard quadratic, ax^2 + bx + c = 0. Go find the solution.
You have A^2 + 8A -105 = 0 (I didn't check for your derivation), so in standard notation this a = 1, b = 8, c = -105. Your "A" is x. Note that only a positive solution for A makes sense.

10. WBahn Moderator

Mar 31, 2012
18,087
4,917
I would recommend stepping back and getting real comfortable with the quadratic equation. You will use this over and over and over again.

If you can manipulate your equation into the form:

(something1)·(something)² + (something2)·(something) + (something3) = 0

then you are in a position to use the quadratic equation to find what (something) can be.

In your case you started out with

C² = A² - 2ABcos(c) + B²

and you want to know A, so A is the (something) and the form of your equation becomes

(1)·(A)² + (-2Bcos(c))·(A) + (B² - C²) = 0

You are now in a position to use the quadratic formula

ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/(2a)

by just substituting in

x = A
a = 1
b = -2B·cos(c)
c = (B² - C²)

If this isn't obvious to you at this point, do yourself a BIG favor and set the trig aside for the moment and go back and work on the algebra.