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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

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."*

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,730

Zmurf,

This formula is easier remembered as

a² = b² + c² - 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

a² = b² + c²,

the Pythogoras Theorem.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

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."*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

Can I have a go at simplifying?

Start : a² = (b sin Θ)² + (c b cos Θ)²

Expand : a² = b² (sin Θ)² + c² 2bc (cos Θ) + b² (cos Θ)²

Combine: a² = b² [(sin Θ)²+(cos Θ)²] + c² 2bc (cos Θ)

Now (sin Θ)²+(cos Θ)² = 1 (a Pythagorean Identity), so:

Finally: a² = b² + c² 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

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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 a² = b² + c² 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.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,730

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

c² = a² + b² - 2ab Cosθ,

you can find the value of c.

Now, use the formula

b² = a² + c² - 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-¹x.

Is that clear?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

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."*

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Sometimes, I really, really want to be older.

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

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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

MathsIsFun wrote:

Can I have a go at simplifying?

Start : a² = (b sin Θ)² + (c b cos Θ)²

Expand : a² = b² (sin Θ)² + c² 2bc (cos Θ) + b² (cos Θ)²

Combine: a² = b² [(sin Θ)²+(cos Θ)²] + c² 2bc (cos Θ)Now (sin Θ)²+(cos Θ)² = 1 (a Pythagorean Identity), so:

Finally: a² = b² + c² 2bc (cos Θ)

(Well done recognising that formula, ganesh)

How did you get to the c² - 2bc (cos θ) + b² (cos θ)² ?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

Well, just by expanding.

If you have (x-y)² you can do this:

(x-y)² = (x-y)(x-y) = x(x-y) - y(x-y) = (x²-xy) - (yx-y²) = x²-xy - yx+y² = x² - 2xy + y² (follow each step carefully)

So, likewise I did:

(c b cos Θ)² = (c b cos Θ)(c b cos Θ) = c(c b cos Θ) - (b cos Θ)(c b cos Θ)

... = c² - bc (cos θ) - bc (cos θ) + b² (cos θ)² = c² - 2bc (cos θ) + b² (cos θ)²

I hope!

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,630

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,730

Zmurf wrote:

How did you get to the c² - 2bc (cos θ) + b² (cos θ)² ?

You can expand the terms as Mathsisfun said.

Or, you may remember the following formulae:-

1. (a+b)² = a² + 2ab + b²

2. (a-b)² = a² - 2ab + b²

3. (a+b)(a-b) = a² - b²

4. (a+b)³ = a³ + 3a²b+3ab²+ b³

5. (a-b)³ = a³ -3a²b +3ab² - b³

6. a³ + b³ = (a+b)(a² -ab +b²)

7. a³ - b³ = (a-b)(a² +ab + b²)

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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