In Rt△OCD, ∠ OCD = 60, OC=6.
∴OD=OC*tan∠OCD=6 root number 3
From the picture, d is on the positive semi-axis of y.
∴D(0,6 root number 3)
∫CB goes through C and D.
∴ ................ (written by himself) straight line CB:Y =- root 3x+6 root 3.
(2) Make B BE⊥OC in E.
P (2,0) can be easily obtained.
Straight line PF divided by p
∴ ................ straight line PF: Y = root 3x-2 root 3.
∫ The straight line PF intersects with the straight line BC in B ..
∴ ............................. B (the radical of 4,2 is 3)
∴BE=Yb=2 Root number 3 (Note: Xb refers to the abscissa value of B, and Yb refers to the ordinate value of B)
AB=Xb=4
∵OC=6,BE⊥OC
∴S trapezoidal OABC =...= 10 root number 3
③qg⊥cp asks after G.
∫ b (4,2 radical number 3), c (6,0)
∴BC=……=4
∫0 < 3t≤4
∴ 0 < t ≤ 4/3
Similarly, 0 < t ≤ 6.
∴ 0 < t ≤ 4/3
At Rt△CQG, ∠ qcg = 60, CQ=3t.
∴ QG = CQ * SIN ∠ QCG = 3/2 root number 3 t
∵CP=OC-OP=6-t,QG⊥CP
∴ s △ pcq =...=-3/4 3 t? +9/2 3ts (it is suggested to write 3/4 3ts * (6-t) when calculating.
AB = 4，BC=4
∴AB=BC
∴∠BCA=∠BAC (standard angle! )
∫AB‖OC
∴∠BAC=∠ACO
∴∠ACO=∠BAC
∴∠ACO = 1∠OCD = 1 * 60 = 30。
In RT△OFC, OC=6, ∠ ACO = 30.
∴OF=sin∠ACO*OC=3
F is the tangent point of circle p.
The radius of circle p is 3.
(1) When ...............
∫S△PCQ =(3+5)0.3s trapezoidal OABC = 3* 10 root number 3 = 15 root number 3.
∴ 15 root number 3 = 3 root number 3 t*(6-t)
The solution is t= 1 or 5 (truncation)
∴t= 1
∴P( 1,0), q [(radical number 3 of 6-3), radical number 3 of 2-3]
Straight line PF divided by p
∴ ................y = root 3x- root 3.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
(If you haven't learned this theorem:
In the plane rectangular coordinate system, in the expression of two mutually perpendicular straight lines,
The coefficient of the first term is reciprocal.
Go straight to the next sign ...)
Let the line passing through o perpendicular PF and m be OM.
The coefficient of its first term is -3 root number 3.
∫ o (0,0)
∴ line Mn: y =-3 root 3x
Pay PF to m
∴ ................................ M (3/4, the root number 3 of -4)
∴ om = ... = the root number of 2 points is 3 < 3.
∴ The straight line PF intersects the circle P.
Let the straight line passing through the q perpendicular PF and n be QN.
Similarly, N (13, 9-3), NB=2-3=2-3.
∴ The straight line PF is tangent to the circle P.
② When? ................
Use (1), t = 3 3 3 points of the square root of 3 (contradiction, abandonment).
To sum up, the straight line PF intersects with the circle p and is tangent to the circle p.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
(logo)
Let the line passing through the O vertical line PF at m be OM, and AO and PF intersect at k..
………………
∴K(0,-radical ranked third)
∴OK= root number 3
In RT△OPK, OP= 1, OK= root number 3.
∴PK=……=2
∴sin∠opk = ok/PK = 2 root number 3.
In RT△OPM, OP= 1, SIN∠OPK = 2 point root number 3.
∴ om = op * sin ∠ opk = root of point 2 3 < 3
∴ The straight line PF is tangent to the circle P.
Let PF and AB intersect at l, and the straight line passing through the vertical line PF of q at n is QN.
∫AB‖OC
∴∠ALP=∠LPB (standard angle! )
∠∠ALP =∠NLB
∠LPB=∠OPK
∴∠NLB=∠OPK
∴sin∠NLB = sin∠opk = 2.
The ordinate of the easily obtained L is 2 and the root number is 3.
L is on PF
∴ .............................. L (3,2 root number 3)
∫ b (4,2 root number 3)
∴…………LB= 1
In RT△LBN, LB= 1, and sin ∠ NLB = the root number 3 of 2.
∴nb = lb * sin∠NLB = 2 root = 2 root = 3.
∴ The straight line PF is tangent to the circle P.
② When? ................
Use (1), t = 3 3 3 points of the square root of 3 (contradiction, abandonment).
To sum up, the straight line PF intersects with the circle p and is tangent to the circle p.
It's over, but one thing, you will find OM‖AC. But I don't know if there is any omission in the calculation process. How many times have you harmed those who copied the answers without thinking! Aren't you ashamed)