King asked in Science & MathematicsMathematics · 8 months ago

# find the angle between the lines xcosα 7+ ysinα=p and xcosβ + ysinβ=q.?

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8 months ago

xcosα + ysinα = p

xcosβ + ysinβ = q

angle between the line = |a₁a₂ + b₁b₂|

cosθ √a²₁ + b²₁ √a²₂ + b²₂

= |cosαcosβ + sinαsinβ| / √cos²α √cos²β + sin²β

= |cos(α-β)| / (1) (1)

cosθ = cos(α-β)

θ = (α-β).

• 8 months ago

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

• 8 months ago

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|