It is known that, as shown in the figure, AB is the chord of ⊙ o, ∠ OAB = 45, C is the point on the superior arc AB, BD ∥ OA intersects CA extension line at point D, connecting BC (1) Verification: BD is the tangent of ⊙ o (2) If AC = 4 √ 3, ∠ cab = 75, find the radius of ⊙ o

It is known that, as shown in the figure, AB is the chord of ⊙ o, ∠ OAB = 45, C is the point on the superior arc AB, BD ∥ OA intersects CA extension line at point D, connecting BC (1) Verification: BD is the tangent of ⊙ o (2) If AC = 4 √ 3, ∠ cab = 75, find the radius of ⊙ o

(1) It is proved that ∵ OA = ob ∵ OAB = ∵ oba = 45 °∵ AOB = 90 ° and BD / / OA ∵ ob ⊥ CD
⊙ BD is the tangent of ⊙ o
(2) For diameter AE and CE, then ∠ ace = 90 °, and ∠ cab = 75 ≠ CAE = 30 ° AC = 4 √ 3,
∴AE=8 ∴R=4
When solving the system of linear equations of three variables {x + 2y-z = 3,2x + y + Z = 5,3x + 4Y + Z = 10, we first eliminate Z and get the system of linear equations of two variables_____________ And then disappear
y. The linear equation of one variable is obtained____________
When solving the system of linear equations of three variables {x + 2y-z = 3,2x + y + Z = 5,3x + 4Y + Z = 10, we first eliminate Z and get the system of linear equations of two variables___ 3x+3y=8,4x+6y=13__________ And then we eliminate y, and we get a linear equation of one variable____ 2x=3________
As shown in the figure, ⊙ O's chord AB = 10, P is a moving point on the superior arc of the chord AB, Tan ∠ APB = 2, (1) if △ APB is a right triangle, find the length of Pb; (2) if △ APB is an isosceles triangle, find the area of △ APB
(1) When △ APB is a right triangle, there are two cases: making diameter ap2 and BPL, connecting PLA and P2b, ∧ P2b = ab △ Tan ∠ APB = 5, PLB = ap2 = 55, so the length of Pb is 5 or 55; (2) when △ APB is an isosceles triangle, there are three cases: ① PA = Pb, ∧ AOH = ∧ APB, ab = 10 ∧ Oh = 52, ∧ OP = 552, pH = 5 + 552 ∧ s △ APB = 25 + 2552; ② Ba = BP, ∧ gab = ∧ APB, take a point P4 on ⊙ o to make bp4 = Ba Let Ag = k ∵ Tan ∠ APB = 2 ∵ BG = 2K, then k = 25 ∵ s △ APB = 40 from Pythagorean theorem; ③ AB = AP is the same as Ba = BP ∵ s △ APB = 40
How to calculate 7x-3y = - 1, 4x-5y = - 17 equations? How to deform 7x-3y = - 1? Y =?
7x-3y=-1 ①
4x-5y=-17②
①*4-②*7=23y-115=0
Y=5
A very simple question, 7x-3y = - 1 becomes 28x-12y = - 4, y = 3
Take the radius of the known circle as 2 and the diameter of the circle AB as the circle M. point C is a moving point on the arc AB of the circle O, connecting AC and BC with the circle m at points D and e respectively,
1. Find the angle c to degree. 2. Find the length of de. 3 if the tan angle ABC = y, ad is greater than DC = x, use the algebraic expression containing x to express y
(1) As shown in the figure: connect OB and OM
Then in RT △ OMB, ∵ ob = 2, MB = 3, ∵ om = 1
∵OM= 12OB,∴∠OBM=30°．
∴∠MOB=60°．
Connect OA. Then ∠ AOB = 120 °
∴∠C= 12∠AOB=60°．
(2) In △ CDE and △ CBA,
∵∠CDE=∠CBA,∠ECD=∠ACB,
∴△CDE∽△CBA,∴ DEAB=DCBC．
Connect BD, then ∠ BDC = ∠ ADB = 90 °
In RT △ BCD, ∵ ∠ BCD = 60 °, ∵ CBD = 30 °, ∵ BC = 2dc
That is, DCAB = 12
∴DE= 12AB= 12×2 3= 3．
(3) Connect AE
∵ AB is the diameter of ⊙ m, ∵ AEB = ∠ AEC = 90 °
From ADDC = x, we can get ad = x &; DC, AC = AD + DC = (x + 1) &; DC
In RT △ ace, ∵ cos ∠ ace = CEAC, sin ∠ ace = aeac,
∴CE=AC•cos∠ACE=（x+1）•DC•cos60°= 12(x+1)•DC；
AE=AC•sin∠ACE=（x+1）•DC•sin60°= 32(x+1)•DC.
From (2), we know BC = 2dc
∴BE=BC-CE= 2DC-12(x+1)•DC=12(3-x)•DC.
In RT △ Abe, Tan ∠ ABC = aebe = 32 (x + 1) &; DC12 (3-x) &; DC = 3 (x + 1) 3-x,
∴ y=3(x+1)3-x（0＜x＜3）．
(1) If we make auxiliary line connection OD, we can prove that the angle Edo is a right angle
Since ∠ C is known to be a right angle, it turns out that △ eco is similar to △ Edo.
There are two conditions: OC = OD, EO = EO. It is proved that ∠ CEO = ∠ OED
E0 parallel Ba
The CEO is equal to B, OED is equal to EDB
∵ - ced is equal to ∠ B ＋ BDE (external angle theorem)
∴∠CEO=∠OED
Therefore, three conditions prove that △ eco and △ Edo are completely expanded
(1) If we make auxiliary line connection OD, we can prove that the angle Edo is a right angle
Since ∠ C is known to be a right angle, it turns out that △ eco is similar to △ Edo.
There are two conditions: OC = OD, EO = EO. It is proved that ∠ CEO = ∠ OED
E0 parallel Ba
The CEO is equal to B, OED is equal to EDB
∵ - ced is equal to ∠ B ＋ BDE (external angle theorem)
∴∠CEO=∠OED
Therefore, three conditions prove that △ eco and △ Edo are congruent
(2) This topic is right angles everywhere anyway. Use Pythagorean theorem to solve it.. Put it away
How to solve 7x-3y + 1 = 0,4x-5y = - 17
7x-3y+1=0 (1),
4x-5y=-17 (2)
(1) 5 - (2) × 3
23x=46
X=2
Substituting (1) to get:
3y=14+1=15
Y=5
Therefore, the solution of the equations is: x = 2; y = 5
We know that the radius of circle O is 2, and take the chord ab of circle O as the diameter to make circle M
Given that the radius of circle O is 2, take the chord ab of circle O as the diameter of circle m, point C is a moving point (not coincident with a and b) on the superior arc AB of circle O, connect AC BC with circle m at points D and e respectively, and connect De, if AB = 2, root sign 3
Finding angle c degree
1 find angle c degree
2 find the length of de
Note that Tan angle ABC = y, ad ratio DC = x (o is less than x, less than 3). In the process of C movement, y is expressed by an algebraic formula containing X
In the triangle AOB, oh is perpendicular to AB, and the perpendicular foot is h. according to the theorem of vertical diameter chord division, ah = BH = 0.5, ab = root 3, and because OA is the radius, OA = 2, in the right triangle AOH, the angle AOH = 60 degrees is obtained. Similarly, the angle boh = 60 degrees, so the angle AOB = 120 degrees, because the angle c is the inner angle of the inferior arc ab
(1) Connect OA, OM
∵ am = BM (M is the center of the circle)
Ψ om ⊥ AB (OM bisector)
∵OA=2, AM=AB/2=√3
‖ om = 1 = OA / 2 (Pythagorean theorem)
∴∠OAM=30, ∠AOM=90-30=60, ∠AOB=60*2=120
The angle of the circle is half of the center angle of the circle
(2) Connect to de, DB
∵ AB is the diameter, ∵ {B}
(1) Connect OA, OM
∵ am = BM (M is the center of the circle)
Ψ om ⊥ AB (OM bisector)
∵OA=2, AM=AB/2=√3
‖ om = 1 = OA / 2 (Pythagorean theorem)
∴∠OAM=30, ∠AOM=90-30=60, ∠AOB=60*2=120
The angle of the circle is half of the center angle of the circle
(2) Connect to de, DB
∵ AB is the diameter, ∵ ADB = 90
∴∠CBD=90-60=30
In △ DEB, by sine theorem:
De / sin30 = 2R (R is the radius of DEB circumcircle ⊙ m) = 2 √ 3
Ψ de = 2 √ 3 * sin30 = √ 3
The third question is y = 1 / 6 √ 3x * x + 1 / 6 √ 3x
{7x-3y+1=0,4x-5y+17=0
When {7x-3y + 1 = 0, 4x-5y + 17 = 0, we deduce {28x-12y + 4 = 0, 28x-35y + 119 = 0, and subtract the two formulas to get 23y = 115, y = 5, and bring in the original formula to get x = 2
The radii of the two concentric circles are 7cm and 2cm respectively. If the chord ab of the big circle is 2 times the root 3, then AB and the small circle
The radii of two concentric circles are 7cm and 2cm respectively. If the chord ab of the big circle is 2 times the root 3, then AB and the small circle ()
A. Separation B, tangency C, intersection D, undetermined
Select b tangent
The distance between the center of a chord and the center of a circle is the distance between the center of a chord. Pythagorean theorem proves that the distance between the center of a chord is 2cm
B
(2x quadratic-5y) (4x-7y) calculation problem
(2x-5y) (4x-7y)
=The quadratic power of 2x (4x-7y) - 5Y) (4x-7y)
=The third power of 8x - the second power of 14x, the second power of y-20xy + 35y
=8x³-14x²y-20xy+35y²
=2x²*4x-2x²*7y-5y*4x+5y*7y =8x³-14x²y-20xy+35y²