hard math question?

Question

please dont solve it for me!! (its not for hw but if i get it right 15 bounus points will be added to my grade and bring it froma c+ to a b+...wich is MUCH BETTER:D..but getting that much bounus points with out working would just not be nice)
just tell me how to solve it if you can

a^2=b^2+c^2-2bc(cos A) solve for b

***^2 means squared
***you will have to use the quatdratic formula
***you will use square roots

if anyone helps me with this you are my hero!!

Answer 1

By rearranging the cosine rule equation above you can get,

b^2 - 2bc(cos A) + (c^2 - a^2) = 0

If we chuck everything into the quadratic formula (where b is like the x you'd normally solve for).

b = [(2c(cos A)) +/- sqrt((2c(cos A))^2 - 4(c^2 - a^2))] / 2

Here +/- is 'plus or minus'.

I figure it'll be bloody hard to read, so you'll have to understand what I've done to copy it down correctly anyway.

Answer 2

hey.
This is the the cosine rule.
i.e you use it to solve triangles
if u have 2 sides and an angle.
this can work for any three angles
a b or c
so if u had a^2 = b^2 + c^2 - 2bc(cos(a))
this can be written in a different way which equates b given that u have angle b .
the formula would be
b^2 = a^2+c^2 - 2ac cos(b)
so b will be square root of wats on the others side.
Hope this helps.

Answer 3

a million+a million=2 besides, somebody with some distance too plenty time on their hands desperate to tutor it... "The evidence starts off from the Peano Postulates, which define the organic numbers N. N is the smallest set relaxing those postulates: P1. a million is in N. P2. If x is in N, then its "successor" x' is in N. P3. there's no x such that x' = a million. P4. If x isn't a million, then there's a y in N such that y' = x. P5. If S is a subset of N, a million is in S, and the implication (x in S => x' in S) holds, then S = N. then you certainly ought to stipulate addition recursively: Def: permit a and b be in N. If b = a million, then define a + b = a' (utilising P1 and P2). If b isn't a million, then permit c' = b, with c in N (utilising P4), and define a + b = (a + c)'. then you certainly ought to stipulate 2: Def: 2 = a million' 2 is in N by potential of P1, P2, and the definition of two. Theorem: a million + a million = 2 evidence: Use the 1st component of the definition of + with a = b = a million. Then a million + a million = a million' = 2 Q.E.D. be conscious: there is yet another formula of the Peano Postulates which replaces a million with 0 in P1, P3, P4, and P5. then you certainly ought to alter the definition of addition to this: Def: permit a and b be in N. If b = 0, then define a + b = a. If b isn't 0, then permit c' = b, with c in N, and define a + b = (a + c)'. you besides mght ought to stipulate a million = 0', and 2 = a million'. Then the evidence of the theorem above is slightly distinctive: evidence: Use the 2d component of the definition of + first: a million + a million = (a million + 0)' Now use the 1st component of the definition of + on the sum in parentheses: a million + a million = (a million)' = a million' = 2 Q.E.D." Wow, he ought to be a real hit with the ladies...!! :)

Answer 4

If the unknown is b write it as a 2nd degree equation
b^2-2c cos A *b +c^2-a^2=0 and apply the formula for thistype of equation


b= (( 2c cos A +-sqrt( 4c^2cos^2A +4(a^2-c^))/2

Answer 5

well first off you will have to isolate your variables or put it into a familiar form.

The quadradic equation form is Ax^2 + Bx + C and in this case your variable (x) is 'b'. Everything else is constant.

So... 0 = b^2 + c^2 -2*b*c*cosA - A^2 - subtract A^2

The first term is easy. It is everything that multiplies b^2 or 1

The second term is harder to see but it is everything that multiplies b so it is -2*c*cosA but not b^2

The third is everything else so is c^2 - a^2

So you have a quadradic where A = 1, B = -2*c*cosA and C = c^2 - a^2 and just to remind you.. the quadradic equation is

x = -b +/- sqrt(b^2 - 4*a*c)

Hope that helps

Answer 6

a^2=b^2+c^2-2bc(cos A)

write it in the following form:

b^2 -2bc(cos A) +(c^2 -a^2)=0

now use the discriminant for the quadratic equation

D = (2c(cos A))^2 -4(c^2 -a^2)

find b1 and b2 roots
b1 =(2c(cos A) - sqrt(D))/2
b2 = (2c(cos A) + sqrt(D))/2