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As shown in the figure, OA, OB and OC are the radii of circle O. if ∠ AOB = 2 ∠ BOC, please judge whether ∠ ACB = 2 ∠ BAC is tenable. Why?
It doesn't work because when C is between a and B, it's obviously not true when C is outside a B
As shown in the figure, known ∠ AOB = α, take OA1 = ob1 on ray OA and ob respectively, connect A1B1, take A2 and B2 on B1a1 and B1B respectively, make b1b2 = b1a2, connect a2b2 According to this law, it is noted that ∠ a2b1b2 = θ 1, ∠ a3b2b3 = θ 2 , an + 1bnbn + 1 = θ n, then （1）θ1= ___ ; （2）θn= ___ .
(1) Let ∠ a1b1o = X,
Then α + 2x = 180 ° x = 180 ° - θ 1,
∴θ1=180°+α
2；
(2) Let ∠ a2b2b1 = y,
Then θ 2 + y = 180 ° ①, θ 1 + 2Y = 180 ° ②,
① × 2 - ② shows that 2 θ 2 - θ 1 = 180 °,
∴θ2=180°+θ1
2；
...
θn=(2n-1)•180°+α
2n．
So the answer is: (1) 180 ° + α
2；（2）θn=(2n-1)•180°+α
2n．
AB = 1 A1B1 = 3 = 1 * 3 a2b2 = 9 = 1 * 3 * 3 = 3 square a3b3 = 27 = 1 * 3 * 3 * () = () a7b7 = ()
A3b3 = 27 = 1 * 3 * 3 * (3) = (cubic of 3)
A7b7 = (the seventh power of 3) = 2187
It is known that A1, A2 and A3 are the three points on the parabola y = 1 / 3x 2, A1B1, a2b2, a3b3, which are perpendicular to the X axis respectively, and the perpendicular feet are B1, B2, B3. The intersection of the straight line a2b2 and a1a3 is at the point C, (1) As shown in Fig. (2), the abscissa of A1, A2 and A3 are 1,2,3, and the line Ca2 = () (2) As shown in Fig. (2), if the parabola y = 1 / 3x? Is changed to the parabola y = 1 / 3x? - x + 1, the abscissa of A1, A2 and A3 are continuous integers, and other conditions remain unchanged, the length of line segment Ca2 is calculated (3) If the parabola y = 1 / 3x? Is changed to the parabola y = ax? + BX + C, A1, A2, A3, the abscissa of the three points is a continuous integer, other conditions remain unchanged, please guess the length of line Ca2 (expressed by a, B, c), and write the answer directly
(1) Method 1:
As shown in the figure, ∠ AOB = 90 ° and OM bisect ∠ AOB. Move the vertex P of the right triangle plate on the ray OM, and the two right angles intersect with OA and ob at points c and D respectively. Is PC equal to PD? Try to explain the reason
PC is equal to PD
Pass point P as PE ⊥ OA at point E, PF ⊥ ob at point F
∵ om bisection ∵ AOB, point P on OM, PE ⊥ OA, PF ⊥ ob,
﹣ PE = PF (the distance from the point on the bisector to both sides of the corner is equal)
And ? AOB = 90 °, PEO = ∠ PFO = 90 °,
The quadrilateral oepf is rectangular,
∴∠EPF=90°，
∴∠EPC+∠CPF=90°，
And ∵ cpd = 90 °,
∴∠CPF+∠FPD=90°，
∴∠EPC=∠FPD=90°-∠CPF．
In △ PCE and △ PDF,
A kind of
∠PEC＝∠PFD
PE＝PF
∠EPC＝∠FPD ，
∴△PCE≌△PDF（ASA），
∴PC=PD．
As shown in Fig. 1, point O is the point on the straight line AB, and the ray OC is made through point O to make ∠ BOC = 120 °. Place the right angle vertex of a right angle triangle plate at point o If the triangle plate rotates around point O, when the distance between point m n and line AB is equal and on both sides of AB, it means that ab bisects Mn
(1) Reason: let the reverse extension line of on be OD, ∵ om bisection ∠ BOC, MOC = ∠ mob, and ∵ om ⊥ on, mod = ∠ mon = 90 degrees, ∵ cod = ∠ Bon, and ? AOD = ∠ BON (equal vertex angle), ∵ cod = ∠ AOD, ∵ od bisection ∠ AOC
As shown in Fig. 1, point O is a point on the line AB, passing through point o as ray OC, so that ∠ BOC = 120 °. Place the right angle vertex of a right angle triangle plate at point O, one side OM is on the ray ob, the other side on is below the line ab (1) Rotate the triangle plate in Fig. 1 anticlockwise around point O to Fig. 2, so that one side OM is in the interior of ∠ BOC, and exactly bisects ∠ BOC (2) Rotate the triangle plate in Fig. 1 around point o at a speed of 6 ° per second in a counter clockwise direction. In the process of rotation, at t second, the straight line on exactly bisects the acute angle ∠ AOC, then the value of T is (write the result directly) (3) Rotate the triangle plate in Figure 1 clockwise around point O to figure 3, so that on is in the interior of ∠ AOC. Please explore the quantitative relationship between ∠ AOM and ∠ NOC, and explain the reasons
(1) Whether the straight line on bisects ∠ AOC
Let the reverse extension line of on be OD,
∵ om bisection ∠ BOC,
∴∠MOC=∠MOB,
And ∵ om ⊥ on,
∴∠MOD=∠MON=90°,
∴∠COD=∠BON,
And ∵ AOD = ∵ BON (equal to vertex angle),
∴∠COD=∠AOD,
ν od bisection ∠ AOC,
That is, straight line on bisection ∠ AOC
（2）∵∠BOC=120°
∴∠AOC=60°,
∴∠BON=∠COD=30°,
That is, when the rotation is 60 ° on bisection ∠ AOC,
From the meaning of the title, 6T = 60 ° or 240 °,
ν t = 10 or 40;
（3）∵∠MON=90°,∠AOC=60°,
∴∠AOM=90°-∠AON、∠NOC=60°-∠AON,
∴∠AOM-∠NOC=（90°-∠AON）-（60°-∠AON）=30°
As shown in the figure, O is a point on the line AB, OC is any ray, OD bisection ∠ BOC, OE bisection ∠ AOC (1) The complementary angles of ∠ AOD and ∠ BOE are pointed out; (2) Try to explain the quantitative relationship between ∠ COD and ∠ Coe
(1) (2) ∠ cod + COE = 12 ∠ AOB = 90 degrees. (Note: because od bisects ∠ BOC, so ∠ cod = 12 ∠ BOC). And OE bisects ∠ AOC, so ∠ COE = 12 ∠ AOC, so ∠ cod + COE = 12 ∠ BOC
As shown in Fig. 1, point O is a point on the line AB, passing through point o as ray OC, so that the angle BOC is 120 degrees. Place the right angle vertex of a right triangle at point O, one side OM is on the ray ob, the other side on is below the line ab (1) Rotate the triangle plate in Figure 1 anticlockwise around point O to figure 2, so that one side OM is inside the angle BOC and exactly bisects the angle BOC. Ask: is the straight line on bisecting the angle AOC? Please explain the reason;
(1) Reason: set the reverse extension line of on as OD, ∵ om bisection ᚛ BOC, ? MOC = ∵ mob, and ∵ om ⊙ on, ? om ⊙ on, ? om ⊙ on, ? mod = ∵ mon = 90 ? om ⊙ on, ? mod = mon = 90 ? cod ᙽ cod ? AOD, ? od bisection ? om bisection whenthe rotation angle is 60 °, on bisection ∠ AOC is equal to or equal to ∠ AOC, According to the meaning of the title, 6T = 60 ° or 240 °, t = 10 or 40; (3) ? mon = 90 °, AOC = 60 °, AOM = 90 ° - ∠ AON, ∠ AOM = (90 ° - ∠ AON) - (60 ° - ∠ AON) = 30 °. Comments: This paper examines the definition of angular bisector, carefully examines the problem and carefully observes the graph, and finds out the relationship between various quantities, which is the key to solve the problem
As shown in the figure, the end point O of ray OC is on the straight line ab. 1. Use a ruler and a compass to make bisectors OM and on2 of ∠ AOC and BOC respectively. Measure the size of ∠ mon with a protractor
Take o as the dot, make a circle, cross OA to D, OC to e, OB to F, and connect de and ef
Find the midpoint of De and EF respectively
Take D and E as the center of the circle, and take the line segment greater than 1 / 2DE as the circle. The two circles intersect at two points g and h to connect GH. GH is the vertical bisector of de. GH and de intersect at I and connect OI. OI is the angular bisector of AOC
and so on!
RELATED INFORMATIONS
- 1. The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD
- 2. As shown in the figure, the side of the cone is cut along OA, and the fan-shaped OAB is expanded into a plane figure (1) What is the relationship between the length of the sector arc AB and the length of the circle at the bottom of the cone? What is the position relationship between point a and point B on the side of the cone? (2) If the angle ∠ AOB = 90 °, what is the relationship between the radius r of the cone bottom circle and the radius r of fan-shaped OAB (i.e. OA or OB)? (3) If point a moves on the side of the cone and then returns to its original position, how should the shortest distance of point a be designed? If R 2 = 0.5, ∠ AOB = 90 °, find the shortest distance of point a
- 3. As shown in the figure, it is known that ∠ AOB = 30 ° and the point P is inside ∠ AOB, Op = 6. If there is a moving point m on OA and a moving point n on ob, then the minimum perimeter of △ PMN is______ .
- 4. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
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- 6. Point O is a point in △ ABC, ∠ ABO = 25 degrees, ∠ ACO = 35 degrees, and the degree of ∠ BOC is 5 degrees less than 2 times of ∠ a In △ ABC, the top angle is ∠ a, the left is ∠ ABC, and the right is ∠ ACB
- 7. In △ ABC, if the bisectors of ∠ a = 50 °, B and C intersect at point O, then the degree of ∠ BOC is () A. 65° B. 100° C. 115° D. 130°
- 8. As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees. If the distance between the point O and the triangle three times is equal, what is the angle AOC? For detailed explanation
- 9. It is known that: as shown in the figure, in △ ABC, the bisector of ∠ B and the bisector of external angle of △ ABC intersect at point D, ∠ a = 90 °
- 10. It is known that the distance between point O and AB and AC is equal, and ob = OC (1) As shown in Figure 1, if point O is on edge BC, it is proved that ab = AC; (2) As shown in Figure 2, if point O is inside △ ABC, it is proved that ab = AC; (3) If point O is outside △ ABC, is ab = AC true? Please draw a picture to show it
- 11. As shown in the figure, the ray OA, ob, OC, OD have common endpoint o, and ∠ AOB = 90 °, COD = 90 °, AOD = 5 4 ﹤ AOC. calculate the degree of ﹤ BOC
- 12. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
- 13. Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
- 14. As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is a point on OC, PD ‖ OA intersects ob at point D, PE ⊥ OA at e, OD = 4cm, then PE ⊥ OA is at point E=______ .
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- 18. As shown in Fig
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