If BC = 6, Ao is equal to the root sign 3, then the circle O is half

If BC = 6, Ao is equal to the root sign 3, then the circle O is half

According to the meaning of the title, the center of a circle should be in an equilateral triangle, BC / 2 = 3, so the vertex angle is generally 30 degrees, the height of the equilateral triangle is 3, the radius of the circle is R ^ 2 = 3 ^ 2 + [(3 roots 3) - root 3] ^ 2, r = root 21

Let's find the number of the product of the vector ABC and the number of the circle ABC

2.5 ah in connection with the geometric meaning of the outer center, the vector BC is decomposed into vector Ba + vector AC, and then the distribution rate vector Ao × vector Ba = module of vector BA × 1 / 2 vector Ba is the same as vector Ao × vector AC = module of vector AC × 1 / 2 vector AC

In the triangle ABC, the angle c is equal to 90 degrees, the angle a is equal to 30, O is a point on AB, Bo is equal to M. the radius of circle O is 1 / 2 When m is taken in what range, the straight line BC and circle O are separated and intersected and tangent?

If the distance between point O and line BC is D, then d = m * cos30 ° = radical 3 * m / 2,
Therefore, when D is equal to the radius of circle O, that is, d = radical 3 * m / 2 = 1 / 2, BC is tangent to the circle, M = 1 / radical 3
When m > 1 / root sign 3, they are separated;
M

In the triangle ABC, ∠ C = 90 ° a = 30 ° o is the point on AB, Bo = a, radius of circle O is root 3, When the line BC and O intersect, find the value range of A When the straight line BC is separated from O, find the value range of A

When OE = √ 3A / 2 < R = √ 3, i.e. a < 2, straight line BC intersects circle O; when OE = √ 3A / 2 = r = √ 3, when OE = √ 3A / 2 = r = √ 3, that is, when OE = √ 3A / 2 = r = √ 3, i.e., a = 2, straight line BC is tangent to circle O; when OE = √ 3A / 2 = r = √ 3, i.e., a = 2, straight line BC is tangent to circle O; when OE = √ 3A / 2 > R =

It is known that: as shown in the figure, in RT △ ABC, ∠ C = 90 ° and the point O is on AB, the circle with o as the center and OA length as the radius intersects with points D and e respectively, and ∠ CBD = ∠ a (1) Judge the position relationship between line BD and ⊙ o, and prove your conclusion; (2) If BC = 2, BD = 5 2. Find ad The value of Ao

（1） The line BD is tangent to ⊙ O
Proof: as shown in Figure 1, connect OD
∵OA=OD，
∴∠A=∠ADO．
∵∠C=90°，
∴∠CBD+∠CDB=90°．
∵ CBD= ∵ A,
∴∠ADO+∠CDB=90°．
∴∠ODB=90°．
The line BD is tangent to ⊙ o
(2) Solution 1: connect de as shown in Figure 1
∵∠C=90°，BC=2，BD=5
Two
∴cos∠CBD＝BC
BD＝4
5．
∵ AE is the diameter of ⊙ o, ᙽ ade = 90 °
∴cosA＝AD
AE．
∵∠CBD=∠A，
∴AD
AE=BC
BD=4
5．
∵AE=2AO，
∴AD
AO=8
5．
Solution 2: as shown in Fig. 2, make oh ⊥ ad at point h through point o
∴AH＝DH＝1
2AD．
∴cosA＝AH
AO
∵∠C=90°，BC=2，BD=5
Two
∴cos∠CBD＝BC
BD＝4
5．
∵∠CBD=∠A，
∴AH
AO=BC
BD=4
5．
∴AD
AO=8
5．

In the RT triangle ABC, the angle c = 90 degrees, the angle B = 30 degrees, O is a point on AB, OA = m, and the radius of circle O is r, When R and M satisfy what relationship, (1) AC intersects circle O, (2) AC is tangent to circle O, (3) AC is separated from circle O, (fast, don 't use cos, sin, etc.),

1: √ 3M < 2R 2: √ 3M = 2R 3: √ 3M > 2R

In RT △ ABC, the angle c = 90 °, angle B = 30 ° o is a point on AB, OA = m, the radius of circle O is r, when R and M satisfy what kind of relationship In RT △ ABC, ∠ C is equal to 90 °, B = 30 ° and O is a point on ab. OA = m, radius of ⊙ o is R. when R and M satisfy what relationship, (1) AC intersects ⊙ o? (2) AC is tangent to ⊙ o? (3) AC is separated from ⊙ o? Draw a picture

As shown in the figure, the tangent of the circle is perpendicular to the radius of the tangent point ᙨ OD

In the RT triangle ABC, the angle ABC = 90 degrees, the point O is on AB, the circle with o as the center and OA length as the radius intersects with points D, e and BD = BC respectively (1) Prove BD as tangent of circle O (2) If BD / CD = 5 / 6, ask for AD / AO value (please help, should use similar, thank you) answer well, there are additional reward points! Thank you

(2) Connecting De, the angle ade = 90 degrees, the angle OED = angle ode = 90 degrees - angle BAC,
BD = BC, angle BDC = angle BCA = 90 degrees - angle BAC,
So the angle OED = angle ode = angle BDC = angle BCA, so angle EOD = angle DBC,
△EOD∽△CBD,(AAA);
OD/DE=BD/DC=5/6,
AO/√(AE²-AD²)=5/6,
AO/√[(2AO)²-AD²]=5/6,
6AO=5√[(2AO)²-AD²]
36AO²=25(4AO²-AD²)
25AD²=64AO²
AD²/AO²=(8/5)²
AD/AO=8/5

In the right triangle ABC, the angle c = 90, the angle B = 30, O is a point on AB, OA = m, the radius of circle O is r, when R and M satisfy what kind of relationship, 1. AC intersects circle o 2. AC is tangent to circle o 3. AC is separated from circle o

When R is equal to three times m of the root sign of two parts, it is tangent; when it is less than three times m of the root sign of two parts, it is separated; when it is more than three times of the root sign of two parts, it intersects

As shown in the figure, ∠ C = 90 °, B = 60 ° in △ ABC, the point O is on AB, Ao = x, and the radius of ⊙ o is 1?

(1) if the circle O is separated from AC, then od is greater than R, i.e. 12x > 1, and the solution is x > 2; (2) if the circle O is tangent to AC, then od is equal to R, i.e. 12x = 1, and the solution is x = 2; (3) if the circle O intersects with AC, then od is equal to R, i.e. 12x = 1, and the solution is x = 2; (3) if circle O intersects with AC, then od is less than R, that is, x = 2; (3) if circle O intersects with AC, then od is less than R, i.e