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

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  • MyRank
    Lv 6
    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).

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  • 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

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  • 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|

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  • 8 months ago

    What does "xcosα 7+" mean? Look up "proofread"

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