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Topic review (newest first)

ganesh
2005-08-02 14:21:31

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)

MathsIsFun
2005-08-02 07:57:35

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!

MathsIsFun
2005-08-02 07:48:30

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!

Zmurf
2005-08-02 07:37:03

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) ?

Roraborealis
2005-08-02 00:44:30

Sometimes, I really, really want to be older.

Zmurf
2005-08-01 21:00:51

yea thanks

ganesh
2005-08-01 20:59:05

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

mathsyperson
2005-08-01 20:54:47

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.

MathsIsFun
2005-08-01 20:42:14

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)

Zmurf
2005-08-01 20:32:36

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?

ganesh
2005-08-01 20:15:20

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

Zmurf
2005-08-01 19:29:47

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

a^2 = (b sin Θ)^2 + (c b cos Θ)^2

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