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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
Draw a triangle randomly, and mark an o point will be more intuitive
（180°-∠BOC）+25°+35°+∠A=180°
∠BOC=2∠A-5°
For the simultaneous equations, the solution is: ∠ a = 65 °
The point O is a point in △ ABC. What is the degree of the angle a = 50 °, OBO = 28 °, ACO = 32 ° and BOC?
In △ ABC
∵∠A=50°
∴∠ABC+∠ACB=180°-70°=130°
∠OBC+∠OCB=130°-28°-32°=70°
In △ OBC
∵∠OBC+∠OCB=70°
∴∠BOC=180°-70°=110°
The degree of ∠ EDC can be obtained as shown in the figure ∠ B = ∠ C, ∠ 1 = ∠ 2, ∠ bad = 40 °
In △ abd, it is known from the external angle properties of triangles that:
(1) ADC = ∠ B + ∠ bad, i.e. ∠ EDC + ∠ 1 = ∠ B + 40 °
In the same way, we get: ∠ 2 = ∠ EDC + ∠ C,
It is known that ∠ 1 = ∠ 2, ∠ B = ∠ C,
∴∠1=∠EDC+∠B，②
② By substituting ①, we can get the following results:
2 ∠ EDC + ∠ B = ∠ B + 40 °, i.e. ∠ EDC = 20 °
The degree of ∠ EDC can be obtained as shown in the figure ∠ B = ∠ C, ∠ 1 = ∠ 2, ∠ bad = 40 °
In △ abd, it is known from the external angle properties of triangles that:
(1) ADC = ∠ B + ∠ bad, i.e. ∠ EDC + ∠ 1 = ∠ B + 40 °
In the same way, we get: ∠ 2 = ∠ EDC + ∠ C,
It is known that ∠ 1 = ∠ 2, ∠ B = ∠ C,
∴∠1=∠EDC+∠B，②
② By substituting ①, we can get the following results:
2 ∠ EDC + ∠ B = ∠ B + 40 °, i.e. ∠ EDC = 20 °
As shown in the figure, in triangle ABC, the angle a = 120 degrees, the bisector of angle ABC and the exterior bisector of angle ACB intersect at point e to find the degree of angle E
∠ABC+∠ACB+∠A=180
So ∠ ABC + ∠ ACB = 60
∠E=180-（∠EBC+∠ECB）
=180-（∠ABC/2+∠ACB/2）
=180-（∠ABC+∠ACB）/2
=180-30
=150
There is a conclusion: e = 90 + ∠ A / 2
You can write it down
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 °
BD bisection ∠ ABC, CBD = 12 ∠ ABC, ? CD bisection △ ABC ? DCE = 12 ∠ ace = 12 (﹤ a + ∠ ABC) = 12 ∠ a + 12 ∠ ABC. In △ BCD, by the external angle property of triangle, ? DCE = ∠ CBD + ∠ d = 12 ∠ ABC + ∠ D, ? 12 ∠ a + 12 ∠ ABC = 12 ﹣ D, ﹤ d = 12 ∠ B
As shown in the figure, △ ABC is the inscribed equilateral triangle of ⊙ o, and the radius of ⊙ o is r (1) Ask for Degree of BC; (2) It is proved that the side length of △ ABC is 3r．
(1) As shown in the figure, connect CO and Bo,
∵ △ ABC is an equilateral triangle,
∴∠BAC=60°；
∴∠BOC=2∠BAC=120°，
Qi
The degree of BC is 120 degrees;
(2) It is proved that, as shown in the figure, OD ⊥ BC is taken as OD ⊥ BC at point D,
∵∠BOC=2∠BAC=120°，BO=CO，
∴∠OCD=30°，
∴DC=COcos30°=
Three
2r，
So BC=
3r．
Given that the equilateral triangle ABC is inscribed on the circle O and the point P is on the arc BC, what is the degree of the angle BPC? This is a third grade problem, my geometry is very poor,
Connecting AP, ∠ BPA = ∠ BCA = 60 degrees, ∠ CPA = ∠ CBA = 60 degrees, ∠ BPC = ∠ CPA + ∠ BPA = 120 degrees
Circle O is the circumscribed circle of triangle ABC, Ao is perpendicular to point F, D is the midpoint of AC arc, and CD arc = 72 degrees. Find the degree of ∠ BAF Circle O is the circumscribed circle of triangle ABC, Ao is perpendicular to point F, D is the midpoint of AC arc, and CD arc = 72 degrees. Find the degree of ∠ BAF
From Ao ⊥ BC, ᚉ AF is the vertical bisector of BC, triangle ABC is isosceles △,
And ab = AC
When CD arc is 72 ° and D is the midpoint of AC arc,
The AC arc is 72 × 2 = 144 °,
The CF arc is 180-144 = 36 °,
Circumferential angle ∠ CAF = ∠ BAF = 36 ÷ 2 = 18 °
It is known that O is a point in the equilateral △ ABC. The ratio of degrees of ∠ AOB, ∠ BOC, ∠ AOC is 6:5:4 To process! It is known that O is a point in the equilateral △ ABC, and the angle ratio of angle AOB, angle BOC and angle AOC is 6:5:4. In the triangle with OA, OB and OC as the sides, the ratio of angles of these three sides is calculated Don't plagiarize. Is there any other solution? Is there any other way besides rotation
The triangle ao'b is obtained by clockwise rotating Δ AOC about point a by 60 ° and connecting OO ' Δ ao'b ≌ AOC ≌ ao'b = ≌ ao'b = ≌ AOC = 96 °, easy to obtain, AOB = 144 °, BOC = 120 °, AOC = 96 °
RELATED INFORMATIONS
- 1. 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°
- 2. 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
- 3. 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 °
- 4. 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
- 5. In the triangle ABC, BC = 2, Sina = 2 / 3, root 2, then what is the maximum product of AB vector and AC vector
- 6. In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process
- 7. In the triangle ABC, a = 60 degrees, AC = 16, area s = 220, root number 3, then BC = what?
- 8. An obtuse triangle ABC, ∠ B is an obtuse angle, where AB = 10, BC = 9, AC = 17, find the height on BC side
- 9. In RT △ ABC, ∠ C = 90 °, ab = 8, AC = 4 times the root sign 3, find the degree of ∠ a, ∠ B
- 10. In the triangle ABC, AB is equal to AC, angle a is equal to 20 degrees, and point D is on AB, so that ad is equal to BC, and the degree of angle BDC is obtained Note: this problem only isosceles triangle and congruent triangle to prove
- 11. Known: as shown in the figure, in the right triangle ABC, the angle c = 90 degrees, the point O is the center of the circle, the length of OA is the radius of the circle, and AC, AB intersect with point D.E, and the angle CBD = angle a.bd is the tangent line of circle O. find: if ad: Ao = 8:5, BC = 2, find the length of BD
- 12. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 13. 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______ .
- 14. 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
- 15. 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
- 16. 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?
- 17. 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
- 18. 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
- 19. 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)
- 20. 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=______ .