In the circle O, AB is the diameter, the chord CD is perpendicular to P, AC = DC = 2 times the root sign 3, and find the length of Op

In the circle O, AB is the diameter, the chord CD is perpendicular to P, AC = DC = 2 times the root sign 3, and find the length of Op

CD=AC,AB⊥CD,CP=CD/2=√3,AP=√（AC^2-CP^2)=3,OP^2=OC^2-PC^2,OP=(AP-AO)=(3-R),CO=R,(3-R)^2+PC^2=R^2,R=2,OP=3-2=1.

As shown in the figure, O is a point on the line AB, ∠ AOC = 1 3 ∠ BOC, OC is the bisector of AOD (1) Find the degree of ∠ cod (2) Judge the relationship between od and AB, and give reasons

(1)  AOC + ∠ BOC = 180 °, AOC = 13 ﹤ BOC,  BOC + ∠ BOC = 180 °, the solution is  BOC = 135 degrees,  AOC = 180 degrees - 135 degrees = 45 degrees,

As shown in the figure, given OD bisection ∠ AOC, ∠ AOB = ∠ cod, ∠ BOC = ∠ AOD, calculate the degree of ∠ AOB

Set option DOC=x,
∵ od bisection ∵ AOC,
∴∠AOD=x，
∵∠AOB=∠COD，∠BOC=∠AOD，
∴∠AOB=x，∠BOC=x，
ν x + X + X + x = 360 ° and x = 90 °,
The degree of AOB is 90 degrees

As shown in the figure, given the angle BOC = 2 angle AOC, OD bisecting angle AOB, and angle cod = 20 degrees, calculate the degree of AOB Don't be too concise,

For the convenience of calculation, let the degree of angle AOB be x degree. ∵ od bisection  AOB (known)  AOD = ∠ BOD = x / 2 (properties of isosceles triangle) ? cod = 20 degrees (known) ? cob = ∠ BOD + ∠ cod = x / 2 + 20, ∠ AOC = ∠ AOD - ∠ cod = x / 2-20 ? cob = 2 ∠ AOC (known) ? X / 2 + 20 = 2 (x /...)

As shown in the figure, the ratio of the degree of AOC to BOC is 5:3, OD bisects ∠ AOB, if ∠ cod = 15 °, calculate the degree of ∠ AOB

Let ∠ AOC = 5x °, then ∠ BOC = 3x °
∴∠AOB=∠AOC+∠BOC=8x°．
∵ od bisection ∵ AOB,
∴∠AOD=1
2∠AOB=4x°．
∵∠COD=∠AOC-∠AOD，
∴5x-4x=15，
∴x=15．
∴∠AOB=8x°=8×15°=120°．
So the answer is: 120 degrees

As shown in the figure, OD is the bisector of AOB, ∠ AOC = 2 ∠ BOC, ∠ cod = 21 ° 30 ′, and calculate the degree of ∠ AOB

Let ∠ BOC = x, then ∠ AOC = 2x, ∵ cod = 21 ° 30 ′,  AOD = 2x-21 ° 30 ′, ∵ BOD = x + 21 ° 30 ′,  od is the bisector of ∠ AOB,

As shown in the figure, ∠ BOC = 2 ∠ AOC, OD bisection ∠ AOB, and ∠ AOC = 40 ° are known to calculate the degree of ∠ cod

∵∠BOC=2∠AOC，∠AOC=40°，
∴∠BOC=2×40°=80°，
∴∠AOB=∠BOC+∠AOC=80°+40°=120°，
∵ od bisection ∵ AOB,
∴∠AOD=1
2∠AOB=1
2×120°=60°，
∴∠COD=∠AOD-∠AOC=60°-40°=20°．

O is a point on the line AB, ∠ BOC = 3 ∠ AOC, OC bisection ∠ AOD (1) Finding the degree of ∠ AOC (2) The relationship between od and ab is inferred, and the reasons are given

∵∠BOC=3∠AOC,∠BOC+∠AOC=180°,
∴3∠AOC+∠AOC=4∠AOC=180°,
∴∠AOC=45°.
（2）OD⊥AB.
Proof: ∵ OC bisection ∠ AOD,
∴∠AOC=∠COD=45°,
∴∠AOD=∠AOC+∠COD=45°+45°=90°,
∴OD⊥AB.
That's it,

Given that ∠ AOB = 25 ° OC ⊥ OA, OD ⊥ ob, then ∠ cod is equal to () A. 25° B. 115° C. 155° D. 25 ° or 155 '

∵OC⊥OA，OD⊥OB，
∴∠AOC=90°，∠BOD=90°．
As shown in Figure 1, ∠ AOD = ∠ BOD - ∠ AOB = 90 ° - 25 ° = 65 °,
Then ∠ cod = ∠ AOD + ∠ AOC = 65 ° + 90 ° = 155 °;
As shown in Figure 2, ∠ BOC = ∠ AOC - ∠ AOB = 90 ° - 25 ° = 65 °,
∠COD=∠BOD-∠BOC=90°-65°=25°．
In conclusion, ∠ cod is equal to 155 ° or 25 °
Therefore, D

Four rays OA, ob, OC and od are caused from point O. if the ratio of ∠ AOB, ∠ BOC, ∠ cod, ∠ doa is 1:2:3:4 ① Finding the degree of ∠ BOC ② If OE bisects ∠ BOC, of, og trisection ∠ cod, calculate ∠ EOG

① Total copies: 1 + 2 + 3 + 4 = 10
∠ BOC = 360 * 2 / 10 = 72 degrees
② ∠ EOG = 72 / 2 + 108 / 3 * 2 = 36 + 72 = 108 degrees
A: not at all