Yahoo Answers: Answers and Comments for Find the angle between the lines xcosα 7+ ysinα=p and xcosβ + ysinβ=q.? [Mathematics]
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From King
enUS
Sat, 25 May 2019 06:58:25 +0000
3
Yahoo Answers: Answers and Comments for Find the angle between the lines xcosα 7+ ysinα=p and xcosβ + ysinβ=q.? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20190525065825AAoLx0D
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From Φ² = Φ+1: Find the angle between the lines cosα x + sinα...
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Sat, 25 May 2019 08:19:55 +0000
Find the angle between the lines cosα x + sinα y = p and cosβ x + sinβ y = q.
cosα x + sinα y = p has an angle of 90° + α {0° ≤ α ≤ 180°}
cosβ x + sinβ y = q has an angle of 90° + β {0° ≤ β ≤ 180°}
The angle between the lines is α  β if this is acute or right, otherwise 180°  α  β

From MyRank: xcosα + ysinα = p
xcosβ + ysinβ = q
angle be...
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Mon, 27 May 2019 08:16:00 +0000
xcosα + ysinα = p
xcosβ + ysinβ = q
angle between the line = a₁a₂ + b₁b₂
cosθ √a²₁ + b²₁ √a²₂ + b²₂
= cosαcosβ + sinαsinβ / √cos²α √cos²β + sin²β
= cos(αβ) / (1) (1)
cosθ = cos(αβ)
θ = (αβ).

From Captain Matticus, LandPiratesInc: I'm assuming you meant:
x * cos(a) + y * ...
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Sat, 25 May 2019 12:37:53 +0000
I'm assuming you meant:
x * cos(a) + y * sin(a) = p
and
x * cos(b) + y * sin(b) = q
y * sin(a) = cos(a) * x + p
y = cot(a) * x + p * csc(a)
y * sin(b) = cos(b) * x + q
y = cot(b) * x + q * csc(b)
So the slopes will be cot(a) and cot(b).
Let's look at something simpler, like y = x. The slope is 1 and that is equal to some tan(t)
tan(t) = 1
t = arctan(1)
So the inverse tangent of the value of the slope will yield an angle for us.
arctan(cot(a))  arctan(cot(b)) = T
T is the angle between them
tan(T) = tan(arctan(cot(a))  arctan(cot(b)))
tan(T) = (tan(arctan(cot(a)))  tan(arctan(cot(b)))) / (1 + tan(arctan(cot(b))) * arctan(cot(a))))
tan(T) = (cot(a)  (cot(b))) / (1 + (cot(b)) * (cot(a)))
tan(T) = (cot(b)  cot(a)) / (1 + cot(a) * cot(b))
tan(T) = (cos(b)/sin(b)  cos(a)/sin(a)) / (1 + cos(a)cos(b) / (sin(a)sin(b)))
tan(T) = ((cos(b)sin(a)  sin(b)cos(a)) / (sin(a)sin(b))) / ((sin(a)sin(b) + cos(a)cos(b)) / (sin(a)sin(b)))
tan(T) = (sin(a)cos(b)  sin(b)cos(a)) / (cos(a)cos(b) + sin(a)sin(b))
tan(T) = sin(a  b) / cos(a  b)
tan(T) = tan(a  b)
T = a  b

From oldschool: What does "xcosα 7+" mean? Look up ...
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Sat, 25 May 2019 07:03:53 +0000
What does "xcosα 7+" mean? Look up "proofread"