If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen

If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen

O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C

As shown in the figure, the diameter of circle O AB and chord CD intersect at point E, and of is perpendicular to CD at F. it is known that AE = 6cm, EB = 2cm, ∠ CEA = 30 ° and find the length of CD

OF=OE*sin(30),AO=(AE+EB)/2=4,OE=AO-EB=2.
In the triangle OFC, OC = Ao = 4, of = 1, and the angle OFC is a right angle, we can get CD = 2 * Radix 15

If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen

O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C

The diameter AB and chord CD of a circle with o as its center intersect at point E, AE = 6, EB = 2, ∠ CEA = 30 ° to find the length of CD

Connect OC and OD and make the vertical line of CD through point O. the vertical foot is f
∵AB=AE+EB=6+2=8
∴OC=OD=AB/2=4
In △ EOF, OE = 2, ∠ OEF = ∠ CEA = 30 °
∴OF=OE/2=1
∴CF=√(4²-1)=√15
CD=2CF=2√15

If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen

O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C

As shown in the figure, AB is the diameter, the chord CD ⊥ AB is at point E, ≁ CDB = 30 ° and the radius of ⊙ o is The length of the string CD is 3 cm______ cm．

∵∠CDB=30°，
∴∠COB=30°×2=60°．
The radius of O is
3cm，
∴CE=
3sin60°=
3 x
Three
2=3
2，
∴CD=3
2×2=3（cm）．

It is known that AB is the chord of ⊙ o, OD ⊥ AB intersects with m ⊙ o at point D, and CB ⊥ AB crosses ad at C (1) Results: ad = DC; (2) If de = 2, CE = 1, find the radius of ⊙ o

(1) ∵ AB is the chord of ⊙ o, radius OD ⊥ AB, CB ⊥ ab,
∴AM=BM，OD∥BC
∴AD=DC．
(2) Connect O and B
The tangent of ⊙ o intersects BC with E,
∴OD⊥DE，
And ∵ OD ⊥ ab,
∴AB∥DE，
∵OD∥BC，OD⊥DE
The quadrilateral mdeb is rectangular,
∵AD=DC，EC=1，DE=2，
∴EC=BE=MD=1，DE=MB=2，
In RT △ BOM, ob2 = om2 + MB2 = (ob-md) 2 + MB2, namely ob2 = (OB-1) 2 + 22,
∴OB=2.5
The radius of ⊙ o is 2.5

It is known that AB is the diameter of ⊙ o, and the chords CD ⊥ AB, e are At a point on AC, the extension lines of AE and DC intersect at point F. it is proved that: ∠ AED = ∠ CEF

Proof: connect ad, as shown in the figure,
∵CD⊥AB，
/ / arc AC = arc ad,
∴∠ADC=∠AED，
∵∠CEF=∠ADC，
∴∠AED=∠CEF．

The chord of the known circle O AB.CD Intersect at point E, the degree of arc AC is 60, the degree of arc BC is 100, what is the degree of angle AEC?

In this case, ∠ AEC = ∠ bad + ∠ CDA. And ∠ bad is the circular angle of arc BD, and ∠ CDA is the circular angle of arc AC. in other words, ∠ bad + ∠ CDA = 1 / 2 (arc BD + arc AC). Therefore, the degree of ∠ AEC is half of the sum of arc AC and arc BD. then the angle AEC = (60 degrees + 100 degrees) / 2 = 80 degrees

If the arc number is known, the arc number is ABCD = 140 Sorry, it's AED = 140 degrees Find ∠ ace

Brother, you have a problem with this topic. If you think about it with your knee, you can see that AC and BD intersect with E, which means that the three points of ace are on the same straight line, where does the angle AEC = 140? (it is a straight line angle, that is, the flat angle is equal to 180 °),