In polar coordinate system, the equation of line L is ρ sin θ = 3, then the distance from point (2, π 6) to line L is () A. 4B. 3C. 2D. 1

In polar coordinate system, the equation of line L is ρ sin θ = 3, then the distance from point (2, π 6) to line L is () A. 4B. 3C. 2D. 1

∵ ρ sin θ = 3, its rectangular coordinate equation is: y = 3, and the rectangular coordinate (3,1) of point (& nbsp; 2 & nbsp;, & nbsp; π 6 & nbsp;). From the distance formula of point to straight line, d = | 3-1 | = 2

Elective course 4-4: coordinate system and parameter equation in polar coordinate system, the curve L: ρ sin2 θ = 2cos θ, passing through point a (5, α) (α is acute angle and Tan α = 34) makes a straight line L parallel to θ = π 4 (ρ∈ R), and l intersects with the curve l at two points B and C respectively. (I) take the pole as the origin, the polar axis as the positive half axis of X axis, and take the same unit length as the polar coordinate to establish a plane rectangular coordinate system and write the curve The general equation of line L and line L; (II) finding the length of | BC |

(I) the Cartesian coordinates of point a are (4,3) and the curve L is ρ 2 & nbsp; Sin2 θ = 2 ρ cos θ, its ordinary equation is: y2 = 2x, because the slope of line L is 1, and it passes through point a (4, 3), so the ordinary equation of line L is: Y-3 = x-4, that is, y = X-1. (II) let B (X1, Y1), C (X2, Y2), by y2 = 2XY = x − 1, we can get x2-4x + 1 = 0, by Weida's theorem we can get X1 + x2 = 4, x1 · x2 = 1, by chord length formula we can get | BC | = 1 + K2 | X1 − x2 | = 26

If the point P (- Sina, COSA) is on the terminal edge of angle B
Then B =?

Analysis
tana=sina/cosa
=-1
So the angle B = 2K π + 3 π / 4

If P (Sina, COSA) is a point on the terminal edge of angle B, then the value of B is zero
The answer is k π + 2 / π - B. I don't understand why it's K π instead of 2K π,

∵ P (Sina, COSA) is a point on the terminal edge of angle β
x=sina,y=cosa,r=|OP|=1
∴sinβ=y/r=cosa=sin(π/2-a)
cosβ=x/r=sina=cos(π/2-a)
When cos β≠ 0, Tan β = Tan (π / 2-A)
∴β=kπ+2/π-a,k∈Z
When cos β = 0, cos (π / 2-A) = 0
∴β=kπ+π/2,π/2-a=mπ+π/2
∴a=mπ
∴β=(k+m)π+π/2-a ,(k,m∈Z)
∧ β = k π + 2 / π - A, K ∈ Z always holds