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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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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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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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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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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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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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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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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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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 ?
"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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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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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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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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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