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

 

2ABcosC - B(2) + C(2) = A(2)

 

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

 

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

But his "C" term is also squared.

 

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

 

I would have values for "C", "B", "cosc" but not "A"
I get stuck when I don't have "A"...

I don't quite understand your question. Are you saying you want to solve explicitly for A, given the others?

 

But his "C" term is also squared.

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.

 

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

 

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.

 

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.

 

i started going through my algebra book, same author as the trig book. there seem to be mistakes in the answers which is aggravating. thanks for the help.

 

i started going through my algebra book, same author as the trig book. there seem to be mistakes in the answers which is aggravating. thanks for the help.

You can count on there always be some level of mistakes, both in the body of the text and in any solutions provided at the end. The solutions are often prepared by grad students that are just getting paid to work problems for the solutions manual, so they don't have a vested interested in making sure that errors are found and corrected. If, however, there are lots of errors, then you should probably find a different book.