As shown in the figure, the radius of ⊙ o is 1cm, and the lengths of strings AB and CD are respectively 2 cm, 1 cm, then the acute angle α between the strings AC and BD=______ Degree

As shown in the figure, the radius of ⊙ o is 1cm, and the lengths of strings AB and CD are respectively 2 cm, 1 cm, then the acute angle α between the strings AC and BD=______ Degree

Connect OA, ob, OC, OD,
∵OA=OB=OC=OD=1，AB=
2，CD=1，
∴OA2+OB2=AB2，
△ AOB is an isosceles right triangle,
Delta cod is an equilateral triangle,
∴∠OAB=∠OBA=45°，∠ODC=∠OCD=60°，
∵∠CDB=∠CAB，∠ODB=∠OBD，
∴α=180°-∠CAB-∠OBA-∠OBD=180°-∠OBA-（∠CDB+∠ODB）=180°-45°-60°=75°．

If the area of the shadow part is 9 π, the length of AB is first?

The shadow area is π * (R ^ 2-r ^ 2) = 9 π, and R ^ 2-r ^ 2 = 9, that is, half of chord AB is 3
AB length is 6

As shown in the figure, AB is the chord of chord circle O, Pb tangent circle O at point B, and op ⊥ OA intersects AB at point C. It is proved that Pb = PC

Connect ob
Because Pb tangent o to point B
So ∠ PbO = 90
That is ∠ oba + ∠ PBA = 90
Because op ⊥ OA
That is, COA = 90
So ∠ a + ∠ ACO = 90
Because OA = ob
So ∠ a = ∠ oba
So ∠ ACO = ∠ PBA
Because ∠ ACO = ∠ PCB
So ∠ PCB = ∠ PBA
So Pb = PC

As shown in the figure, PA tangent circle O at point a, chord ab ⊥ OP, chord perpendicular foot m, ab = 4, OM = 1, then the length of PA

Link OA
Because PA is tangent to point a, PA ⊥ OA
And ab ⊥ OP, the chord perpendicular foot is m, and ab = 4
According to the properties of the circle, am = 1 / 2 * AB = 2
In RT △ OAM, OM = 1. According to Pythagorean theorem, OA = √ 5
And ∠ OAP = ∠ OMA = 90 ° and ∠ AOP is the common angle of RT △ OAM and RT △ POM
So RT △ OAM ∽ RT △ OPA (AA)
Then am / PA = OM / OA
So PA = am * OA / OM
=2*√5/1
=2√5

Point P is a point in a circle with o as the center of radius 5, and op = 3. Among all chords with o as the center of circle passing through point P, the number of chords with integer chord length is?

The longest center of the circle, equal to the diameter, equals 10
The shortest, perpendicular to the longest, is equal to 8
So there should be three, which are 8, 9, 10

Given that point P is a point in circle O with radius 5, tangent OP = 4, among all chords of circle O passing through point P, the number of chords with integral chord length is () A 6 B 5 C 4 D 2

B
Because the longest chord passing through point P is diameter with length of 10, and the shortest chord is the one perpendicular to op with length of 6, so all chords passing through point P are between 6 and 10, and there are 6, 7, 8, 9, 10, totally 5

If there is a point P in ⊙ o with radius of 13, Op = 12, then the number of chords passing through point P and of integer length is () A. 2 B. 17 C. 32 D. 34

The longest chord passing through point P is diameter and length is 26;
The shortest chord passing through P is perpendicular to Op
Connect OA, in right angle △ OAP, AP=
OA2−OP2=
132−122=5，
Then AB = 2AP = 10
Then the range of chord length passing through point P is: greater than or equal to 10 and less than or equal to 26,
In these 17 numbers, there is only one chord except 10 and 17, and the other values have two, then the number of strings is: 2 + 2 (17-2) = 32
Therefore, C

Point P is a point in ⊙ o with radius of 5, and op = 4. Among all ⊙ o chords passing through point P, the number of chords with integer chord length is () A. 8 B. 7 C. 5 D. 3

As shown in the figure, AB is the diameter, OA = 5, Op = 4, passing through point P as CD ⊥ AB, intersecting the two points of point C and D. according to the vertical diameter theorem, point P is the midpoint of CD, obtained by Pythagorean theorem, PC = 3, CD = 6, then CD is the shortest chord passing through point P, with length of 6; ab is the longest chord over P, with length of 10

As shown in the figure, AC is known to be the diameter of circle O, PA ⊥ AC, connecting OP, chord CB parallel OP, straight line Pb intersecting line AC with D, BD = 2PA Find sin ∠ OPA

Join OP, ab. intersect point E
∵ op ∥ BC, ab ⊥ BC, ∵ AOP = ∠ ACB ≁ Bao = ∠ OPA, ∵ AEO = ∠ ABC, i.e., Op ⊥ AB,

As shown in the figure: given that AC is the diameter PA of circle O, vertical AC, connecting OP, chord CB parallel OP, straight line Pb intersecting straight line AC at D, BD = 2PA, it is proved that Pb is tangent of circle o

∵cb//op
∴∠aop=∠acb
∵ ob = OC (BC is the string)
∴∠acb=∠obc
∵cb//op
So ∠ OBC = BOP
∴∠aop=∠acb=∠obc=∠bop
There are ob = OA, Op = Op
∴△aop≌△bop
∴∠obp=∠oap=90°
So Pb ⊥ ob, Pb is tangent of circle o
[without BD = 2PA condition]