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Ray OA, ob, OC, OD have a common endpoint o, and OA is perpendicular to ob, OC is perpendicular to OD, ∠ AOD = 5 / 4 ∠ AOC, calculate the degree of ∠ BOC ∠ AOD = 5 / 4 ∠ AOC! No, ∠ AOD = 5 / 4 ∠ BOC
The solution is as follows: ∵ AOD = 5 / 4 AOD = 5 / 4 AOC, AOD = 5x, AOC = 4x
Four rays OA, ob, OC, od are drawn from point O. if ∠ AOB: ∠ BOC: ∠ cod: ∠ DOA = 1:2:3:4, then the degree of these four angles is ∠ AOB=______ ，∠BOC=______ ，∠COD=______ ，∠DOA=______ .
AOB: BOC: ∠ cod: ∠ DOA = 1:2:3:4, AOD ≠ AOB + ∠ BOC + ∠ cod, i.e. ? AOD + ∠ AOB + ∠ BOC + ∠ cod = 360 °, let ｛ AOB = x °, BOC = 2x °, COD = 3x °, then x + 2x + 3x + 4x = 360, x = 36, ﹤ AOB = 36 ° and ∠ BOC = 72 °
As shown in Fig. 3, four rays OA, ob, OC, od are successively drawn from point O, if angle AOB; angle BOC; angle cod;; angle DOA = 1:3:5:6, calculate angle AOB, angle BOC,
Let ∠ AOB be a, then from the meaning of the title, a + 3A + 5A + 6A = 360 ° and a = 24 °, so ∠ AOB = 24 ° and ∠ BOC = 72 °
As shown in the figure, four rays (OA, ob, OC, OD) are emitted from point O. it is known that ∠ AOC = ∠ BOD = 90 ° 1. If ∠ BOC = 55 °, calculate the size of ∠ AOB and ∠ cod? 1. If ∠ BOC = 55 °, calculate the size of ∠ AOB and ∠ cod? 2. If ∠ BOC = α °, the size of ∠ AOB and ∠ cod?
1.∠AOB=∠AOC-∠BOC=35°
∠COD=∠BOD-∠BOC=35°
2.∠AOB=∠AOC-∠BOC=90°-α°
∠COD=∠BOD-∠BOC=90°-α°
As shown in the figure, PA is the tangent line of circle O, a is the tangent point of circle O, and PBC is the secant of circle o BC= Three 2, then Pb BC= ___ .
∵PA
BC=
Three
2, ♀ PA can be set=
3x，BC=2x
∵ PA is the tangent of circle O, and a is the tangent point,
ν pa2 = PC × Pb, i.e（
3x）2=PB（PB+2x）
From the solution, Pb = x, combined with BC = 2x, Pb is obtained
BC=1
Two
So the answer is: 1
Two
As shown in the figure, PBC is the secant of circle O. if PD = 1, ad = 2, be = 3, then AC =? 0 is the secant of circle O. if PD = 1, ad = 2, be = 3, then AC =? AC = 0?
∵DE∥AC,PD=1,AD=2,
∴DB/AC=PD/PA=1/3,
∴AC=3DB,
PA tangent circle O at point a, according to the cutting line theorem, Da ^ 2 = DB * De, be = 3,
ν 4 = dB (DB + 3), DB = 1
∴AC=3.
As shown in the figure, AB is the diameter of ⊙ o, PA cuts ⊙ o in a, Op intersects ⊙ o in C, and connects BC. If ∵ P = 30 °, find the degree of ⊙ B
⊙ PA cuts ⊙ o to a, AB is the diameter of ⊙ o,
∴∠PAO=90°，
∵∠P=30°，
∴∠AOP=60°，
∴∠B=1
2∠AOP=30°．
PA is tangent of circle O, a is tangent point, secant PBC crosses center O, PA = 4, Pb = 2.1 find the length of BC and AB, if the bisector of BAC intersects BC and circle O respectively PA is the tangent of circle O, a is the tangent point, and the secant PBC passes through the center O, PA = 4, Pb = 2 Find the length of BC and ab 2 if the bisector of the angle BAC intersects points D and e respectively with BC and O, find the length of AE
Try to do it
1. According to the cutting line theorem, there are
PA²=PB*PC
16=2PC
PC=8
BC=PC-PB=8-2=6
Radius of circle = 1 / 2BC = 3
Because △ PAB ∽ PCA
So PA / PC = AB / AC
4/8=AB/AC
AB/AC=1/2
Let AB = k, AC = 2K
Pythagorean theorem
AB²+AC²=BC²
k²+4k²=36
k=6/√5
AB=k=6√5/5,AC=12/√5
2. Ad is the bisector of the angle BAC
Then AB / AC = BD / CD
1/2=BD/CD
So BD = 1 / 3bC = 2, CD = 2 / 3bC = 4
∠ bad = 45 degrees in triangle bad
Sine theorem
BD/sin45=AD/sin∠ABD
In the right triangle BAC, sin ∠ abd = AC / BC = (12 / √ 5) / 6 = 2 / √ 5
2/(√2/2)=AD/(2/√5)
AD=4√10/5
AE and BC intersect at point D
AD*DE=BD*DC
4√10/5*DE=2*4
DE=√10
AE=4√10/5+√10=9√10/5
reference resources
It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
It is proved that: square ABCD, AB = CD, bad = ∠ CDA = 90 °, ? pad = PDA = 15 °, PA = PD, ∠ PAB = ∠ PDC = 75 °, make △ DGC and △ ADP congruent in the square, ? DP = DG, ∠ ADP = ∠ GDC = ∠ DAP = ∠ DCG = 15 °, ? PDG = 90 ° - 15 ° = 60 °, Δ PDG
It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
It is proved that: square ABCD, AB = CD, bad = ∠ CDA = 90 °, ? pad = PDA = 15 °, PA = PD, ∠ PAB = ∠ PDC = 75 °, make △ DGC and △ ADP congruent in the square, ? DP = DG, ∠ ADP = ∠ GDC = ∠ DAP = ∠ DCG = 15 °, ? PDG = 90 ° - 15 ° = 60 °, Δ PDG
RELATED INFORMATIONS
- 1. It is known that: as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. Verification: △ PBC is an equilateral triangle
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- 7. 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?
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- 9. 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
- 10. 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______ .
- 11. As shown in Fig
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- 13. Mathematical problem: introduce two tangent lines PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1 1, introduce two tangent PA, Pb from the moving point P to the circle x ^ 2 + y ^ 2 = 1, and the tangent points are a, B, and ∠ APB = 60 degrees, then the trajectory equation of the moving point P is________ 2. Let P (x, y) be a moving point on the circle x ^ 2 + (Y-1) ^ 2 = 1. If the inequality x + y + C > 0 holds, then the value range of C_____________ 3. We know that the equation of the line where the edge ab of the equilateral △ ABC is located is x √ 3 + y = 0, and the coordinates of point C are (1, √ 3) 4. It is known that the circle M: x ^ 2 + (Y-2) ^ 2 = 1, q is the moving point on the x-axis, QA and QB tangent the circle m to two points a and B respectively (1) (liberal arts) if | ab | = 4 √ 2 / 3, find the equation of line MQ (Science) prove that the straight line AB always crosses a fixed point (2) Find the locus equation of midpoint P of moving string ab It's better to analyze it
- 14. As shown in the figure, lead the tangent PA, Pb of the circle through a point p outside ⊙ o with radius of 6cm, connect Po, intersect ⊙ o at F, make tangent of ⊙ o through F, intersect PA and Pb at D and e respectively, if Po = 10cm, ﹤ APB = 40 ° Find: (1) the circumference of △ ped; (2) the degree of ∠ doe
- 15. Pass through the plane α of the triangle ABC and a point p outside the plane α of △ ABC, make Po ⊥ α, and connect PA, Pb, PC with the perpendicular foot of O 2. If PA = Pb = PC, then o is △ ABC__ heart
- 16. As shown in the graph, P is a point in the equilateral △ ABC, PA = 6, Pb = 8, PC = 10, if the point P 'is a point outside △ ABC, but △ p'ab is congruent with △ PAC The distance between point P and point P 'and the degree of ∠ APB
- 17. As shown in the figure, BC is the diameter of ⊙ o, P is a point on ⊙ o, a is the midpoint of arc BC, ad ⊥ BC, perpendicular foot is D, Pb intersects AD and AC respectively at points e, F, AE and be? Why? It stops after 20 minutes
- 18. It is known that P is a point in the plane of triangle ABC. If vector CB = γ vector PA + vector Pb, γ belongs to R, then point P must be in the plane A. Inside the triangle ABC B. On the line where the AC side is located C. On the line where AB side is located D. On the line where the BC side is located Why?
- 19. Plane PAC ⊥ plane ABC, PA = Pb = PC, verify ab ⊥ BC
- 20. I come from China. Or I'm come from China?