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#1 2005-08-01 19:29:47

Zmurf
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Has anyone seen this formula before?

Has anyone seen this formula before? Or one very similar?

a^2 = (b sin Θ)^2 + (c b cos Θ)^2


"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."

#2 2005-08-01 20:15:20

ganesh
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Re: Has anyone seen this formula before?

Zmurf,
This formula is easier remembered as

a = b + c - 2bc cosθ
On simplification of the equation given by you, this is what is obtained.
a,b, and c are the three sides of a  triangle.
If the triangle is rightangled and θ is 90 degrees or pi/2 radians,
Cos θ = 0,
which gives us
a = b + c,
the Pythogoras Theorem. smile


Character is who you are when no one is looking.

#3 2005-08-01 20:32:36

Zmurf
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Re: Has anyone seen this formula before?

Oh ok. But what formula would I use if i want to find the angle oposite to side 'b' if i'm only given the lengths of sides 'a' and 'b' and the angle between them?


"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."

#4 2005-08-01 20:42:14

MathsIsFun
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Re: Has anyone seen this formula before?

Can I have a go at simplifying?

Start      : a = (b sin Θ) + (c b cos Θ)
Expand  : a = b (sin Θ) + c 2bc (cos Θ) + b (cos Θ)
Combine: a = b [(sin Θ)+(cos Θ)] + c 2bc (cos Θ)

Now (sin Θ)+(cos Θ) = 1 (a Pythagorean Identity), so:

Finally: a = b + c 2bc (cos Θ)

(Well done recognising that formula, ganesh)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#5 2005-08-01 20:54:47

mathsyperson
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Re: Has anyone seen this formula before?

Here:                       A                             
                               /\
                             /    \
                        c /        \  b
                         /            \
                     B ------------- C
                                a

If you know two sides and the angle between them, then in this triangle that would be angle A and sides b and c.
You can't find angle B directly from this, but you can use a = b + c 2bc (cos A) to find side a.
Knowing this, you can use sin A/a=sin B/b to find angle B.


Why did the vector cross the road?
It wanted to be normal.

#6 2005-08-01 20:59:05

ganesh
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Re: Has anyone seen this formula before?

Zmurf wrote:

Oh ok. But what formula would I use if i want to find the angle oposite to side 'b' if i'm only given the lengths of sides 'a' and 'b' and the angle between them?

With the help of the formula
c = a + b - 2ab Cosθ,
you can find the value of c.
Now, use the formula
b = a + c - 2ac Cosθ.
You know the value of a,b, and c.
You would get
Cosθ = x (some value)
The angle opposite to side 'b' would then be
θ = Cos-x.
Is that clear? smile


Character is who you are when no one is looking.

#7 2005-08-01 21:00:51

Zmurf
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Re: Has anyone seen this formula before?

yea thanks


"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."

#8 2005-08-02 00:44:30

Roraborealis
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Re: Has anyone seen this formula before?

Sometimes, I really, really want to be older.


School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

#9 2005-08-02 07:37:03

Zmurf
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Re: Has anyone seen this formula before?

MathsIsFun wrote:

Can I have a go at simplifying?

Start      : a = (b sin Θ) + (c b cos Θ)
Expand  : a = b (sin Θ) + c 2bc (cos Θ) + b (cos Θ)
Combine: a = b [(sin Θ)+(cos Θ)] + c 2bc (cos Θ)

Now (sin Θ)+(cos Θ) = 1 (a Pythagorean Identity), so:

Finally: a = b + c 2bc (cos Θ)

(Well done recognising that formula, ganesh)

How did you get to the c - 2bc (cos θ) + b (cos θ) ?


"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."

#10 2005-08-02 07:48:30

MathsIsFun
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Re: Has anyone seen this formula before?

Well, just by expanding.

If you have (x-y) you can do this:

(x-y) = (x-y)(x-y) = x(x-y) - y(x-y) = (x-xy) - (yx-y) = x-xy - yx+y = x - 2xy + y     (follow each step carefully)

So, likewise I did:

(c b cos Θ) = (c b cos Θ)(c b cos Θ) = c(c b cos Θ) - (b cos Θ)(c b cos Θ)
... =  c - bc (cos θ) - bc (cos θ) + b (cos θ) = c - 2bc (cos θ) + b (cos θ)

I hope!


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#11 2005-08-02 07:57:35

MathsIsFun
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Re: Has anyone seen this formula before?

Roraborealis wrote:

Sometimes, I really, really want to be older.

You are just on the edge of learning this, I think. For the moment don't be "overawed" by it. In a few years you will say "ah, obviously!"

It is just one or two levels further into "The Game".

Please ask questions yourself!


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#12 2005-08-02 14:21:31

ganesh
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Re: Has anyone seen this formula before?

Zmurf wrote:

How did you get to the c - 2bc (cos θ) + b (cos θ) ?

You can expand the terms as Mathsisfun said.
Or, you may remember the following formulae:-

1. (a+b) = a + 2ab + b
2. (a-b) = a - 2ab + b
3. (a+b)(a-b) = a - b
4. (a+b) = a + 3ab+3ab+ b
5. (a-b) = a -3ab +3ab - b
6. a + b = (a+b)(a -ab +b)
7. a - b = (a-b)(a +ab + b)


Character is who you are when no one is looking.

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