As shown in the figure, given Ao ⊥ OC, ob ⊥ OD, ∠ AOB = 142 °, calculate the degree of ∠ cod

As shown in the figure, given Ao ⊥ OC, ob ⊥ OD, ∠ AOB = 142 °, calculate the degree of ∠ cod

∵AO⊥OC，OB⊥OD，
∴∠AOC=∠DOB=90°，
∵∠AOB=142°，
∴∠BOC=142°-90°=52°，
∴∠COD=90°-52°=38°．

As shown in the figure, given Ao ⊥ OC, ob ⊥ OD, ∠ AOB = 142 °, calculate the degree of ∠ cod

∵AO⊥OC，OB⊥OD，
∴∠AOC=∠DOB=90°，
∵∠AOB=142°，
∴∠BOC=142°-90°=52°，
∴∠COD=90°-52°=38°．

As shown in the figure, OA ⊥ OD, OC ⊥ ob, ∠ AOB = 2 ∠ cod, calculate the degree of ∠ AOB Come on, it's due tomorrow B D C O A

Let ∠ cod = x, then ∠ AOC = 90-x, ∠ BOD = 90-x
∴∠AOB=90-x+90-x+x=180-x
∵∠AOB=2∠COD
∴180-x=2x
3x=180
x=60
∴∠AOB =180-60=120°

As shown in Figure 12, in △ AOB, OA = ob, ∠ AOB = 90 ° in △ cod, OC = OD, ∠ cod = 90 ° first, overlap △ AOB with right angle vertex o of △ cod. When △ cod is rotated clockwise around point O, what is the size relationship between the connecting line AC and BD of the other two vertices? Please guess and explain your conclusion

The explanation is as follows:
∵ AO=OB,CO=OD
And ∠ AOB = ∠ cod
∴ ∠AOC=∠BOD
The △ AOC and △ BOD congruence
∴ AC=BD
In other words, no matter how the △ cod rotates around the point O, AC and BD are equal.
The conclusion remains unchanged when ∠ AOB = ∠ cod = 90 ° is replaced by ∠ AOB = ∠ cod = 60 °.

Take any point O in the straight line AB, and pass through the point o as the ray OC, OD, so that ∠ cod = 90 ° when ∠ AOC = 30 °, the degree of ∠ BOD is

60 ° or 120 '

It is known that: ∠ AOB = 100 ° and two straight lines OC and od are drawn from the vertex o of the angle, so that ∠ AOC = 30 ° and ∠ BOD = 40 ° Find the degree of ∠ cod

① As shown in the figure:
∠COD=∠AOB-∠AOC-∠BOD=30°；
② As shown in the figure:
∠COD=∠AOB+∠AOC-∠BOD=90°；
③ As shown in the figure:
∠COD=∠AOB+∠AOC+∠BOD=170°；
④ As shown in the figure:
∠COD=∠AOB-∠AOC+∠BOD=110°．

Take any point O in the line AB, and pass through the point O to make OC perpendicular to OD. When the angle AOC = 30 degrees, calculate the degree of angle BOD

① When OC and od are on the side of AB, ∵ OC ⊥ OD, ∵ cod = 90 °, AOC = 30 °, BOD = 180 ° - ∠ cod - ∠ AOC = 60 °; ② when OC and od are on both sides of AB, ? OC ⊥ OD, ∵ AOC = 30 °, ∵ AOD = 60 °, ∵ BOD = 180 ° - ﹤ AOD = 120 °

(1) Take any point o on the line AB and pass through the point o as the ray OC and OD, so that ∠ cod = 90 ° and ∠ AOC = 30 ° and calculate the degree of ∠ BOD

Angle cod = 90 °, angle COA = 30 °, angle AOD = 90 ° - 30 ° = 60 °, angle BOD = 180 ° - 60 ° = 120 °

Take any point o on the line AB, and pass through the point O to make OC ⊥ OD. When ∠ AOC = 30 °, the degree of ∠ BOD is 0______ .

When OC and od are on the same side of line AB, as shown in the figure:
∵OC⊥OD，∠AOC=30°；
∴∠BOD=180°-∠COD-∠AOC=180°-90°-30°=60°；
When OC and od are on the opposite side of line AB, as shown in the figure:
∵OC⊥OD，∠AOC=30°；
∴∠BOD=180°-∠AOD=180°-（∠DOC-∠AOC）=180°-（90°-30°）=120°．

AB is the chord of circle O. the radii OC and OD intersect AB at points e, F and AE respectively= BF.OE What does it have to do with the size of of of? Why?

Make om ⊥ AB by O, and m by vertical foot
∴AM=BM
∵AE=BF
∴OE=OF
∵oM⊥AB
∴OE=OF