![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
0 is a point in the triangle ABC, and the distance from O to the three sides is equal. Given the angle a = 70 °, calculate the degree of BOC
125°
When Ao is connected and BC is extended to D, the angle BOC = angular BOD + angular cod
Angle BOD = angle Bao + angle oba (angle BOD is the outer angle of triangle AOB, Bo bisection angle b)
Angle cod = angle CaO + angle OCA (angle cod is the outer angle of triangle AOC, CO bisection angle c)
Angle BOC = angle a + 1 / 2 (180 ° - angle a) = 90 ° + angle A / 2 = 125 °
If O is the outer center of the triangle ABC and the angle BOC = 140 degrees, find the degree of angle A
Because o is the outer center of the triangle ABC, so OA = ob = OC, so ∠ OAC = ∠ OCA, ∠ OAB = ∠ oba, ∠ OBC = ∠ OCB because ∠ BOC = 140 ° so ∠ OBC = ∠ OCB = 20 ° and ∠ OAC + ∠ OCA + ∠ oba + ∠ OBC + ∠ OCB = 180 ° so ∠ OAC + ∠ OCB = 180 degrees
O is the outer center of the triangle ABC, ∠ BOC = 80 degrees, ∠ a =?
If O is in the interior of △ ABC, then ∠ a = 40 degrees
If O is outside △ ABC, then ∠ a = 140 °
If O is the outer center of △ ABC and ∠ BOC = 60 °, then ∠ BAC=______ .
Because ∠ BOC is
The angle of the center of the circle to which BC is directed, ∠ BAC is
The circular angle of BC,
Therefore, there are two cases: ① ∠ BAC = 1
2 ∠ BOC = 30 degrees, ② ∠ BAC = 1
2（360°-∠BOC）=150°．
As shown in the figure, △ ABC is inscribed in ⊙ o, AC = 1, ABC = 45 ° then ⊙ o radius=______ .
Make diameter ad through point a and connect DC, as shown in the figure,
∵ ad is the diameter,
∴∠ACD=90°，
And ? ADC = ∠ ABC = 45 °,
The △ ADC is an isosceles right triangle,
∴AD=
2AC，
And AC = 1,
∴AD=
2，
So the radius of ⊙ o is zero
Two
2．
In the triangle ABC, the angle a = 120 degrees, ab = 3, AC = 2, find BC and SINB
Make the vertical line of AB through point C and the extension line of BA at D
Angle CAD = 60 degrees
In triangle ACD: ad = 1, CD = radical 3
In triangle BCD, BC ^ 2 = CD ^ 2 + BD ^ 2, BC = root 19, SINB = root 3 / root 19
In the triangle ABC, the angle a = 120 degrees, ab = 4, AC = 2, then the value of SINB Thank you for the details
Junior high school solution:
The high CD of AB side is made by C,
∵ a = 120 degrees, CAD = 60 degrees, ACD = 30 degrees
∵AC=2,∴AD=1,CD=√3 ；
∵AB=4,∴BD=5,BC=2√7
Then SINB = CD / BC = √ 3 / 2 √ 7 = √ 21 / 14
As shown in the figure, in the triangle ABC, the angle BAC = 105 degrees, the angle B = 45 degrees, and ab = 2 roots
Because the angle BAC = 105 degrees, the angle B = 45 degrees, so ∠ C = 30 degrees. And because AB = 2 root sign 2, according to the sine theorem, AC = 4, BC = 2 + 2, root sign 3
In the triangle ABC, if a = radical 2, B = 2, SINB + CoSb = radical 2, then ∠ a =? RT
SINB + CoSb = square root 2, sin2b = 1, 2b = 90, B = 45
Using the sine theorem a / Sina = B / SINB
Root 2 / Sina = 2 / (root 2 / 2)
Sina = 1 / 2 a = 30 or 150 (impossible, omit)
As shown in the figure, it is known that in △ ABC, ∠ B = 90 ° o is a point on ab. if a circle with o as its center and ob as its radius intersects with point E, and with AC tangent to point D, ad = 2, AE = 1, then the length of ab is______ , the length of the CD is______ .
∵ ad is ⊙ o is tangent, ᙽ ad2 = AE · ab. ? ad = 2, AE = 1. ? 22 = 1 × AB, ab = 4. ? B = 90 °, ac2 = ab · BC. (2 + CD) 2 = 42 + BC2, ∵ B = 90 °, ab is the diameter of ⊙ o, ? CB is the tangent of ⊙ o. ᙽ CD = CB, (2 + CD) 2 = 42 + CD2, CD = 3
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ∠ B = 90 ° o is a point on AB, and a circle with o as its center and ob as its radius intersects with point e at point E
- 2. It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°
- 3. As shown in the figure, take the diameter BC of ⊙ o as an equilateral ⊙ ABC, AB and AC intersect ⊙ o at D and e. it is proved that BD = de = EC
- 4. As shown in the figure, ad is the center line of △ ABC, ∠ ADC = 45 ° and BC = 2cm. Fold △ ACD along ad so that point C falls at the position of E, then be=______ cm．
- 5. As shown in the figure, in the RT triangle ABC, ∠ = 90 ° ad bisect ∠ BAC, and ∠ B = 3 ∠ bad, find the degree of ∠ ADC
- 6. As shown in the figure △ ABC, ab = AC point D is a point on CB extension line ∠ ADB = 60 ° point E is a point on AD and de = dB, it is proved that AE = be + BC Pictures can be drawn according to the theme!
- 7. As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, points E and F are on AB and AC respectively, and AE = AF, it is proved that de = DF
- 8. As shown in the figure, translate the right triangle ABC along the BC direction to the position of the right triangle def, ab = 8, be = 5, Ge = 5
- 9. In the triangle ABC, the angle c = 90 degrees, D is a point on the edge AC, and ad = 1, angle abd = 45 degrees. Let the angle CBD = a, BC = X. when Tana = 3 / 5, find x =? When x = 0.16, find Tana =?
- 10. As shown in the figure, in △ ABC, ∠ ACB = 90 °, point E is the midpoint of AB, connecting CE, passing through point E as ed ⊥ BC at point D, taking a point F on the extension line of De to make AF = CE
- 11. It is known that: point I is the inner heart of the triangle ABC, point O is the outer center of the triangle ABC, ∠ BOC = 140 ° to find ∠ BIC? Why is ∠ a = 70 ° or 110 ° more detailed?
- 12. As shown in Figure 1, in triangle ABC, OB and OC are bisectors of angle ABC and angle ACB respectively. If angle a = x degrees, what is the degree of angle BOC
- 13. As shown in the figure, BC is the chord of circle O, OD ⊥ BC is at point E, intersection arc BC is at point D, and point a is a moving point on the major arc BMC (not coincident with B and C). BC = 4, radical 3, ed = 2 are known (1) Find the radius of circle O (2) Find the degree of ∠ a (3) Find the position of point a when the shadow part takes the maximum value I can't figure it out
- 14. In RT △ ABC, ∠ C = 90 ° cab = 60 ° and the distance between D and ab is 3.8cm, then how many centimeters is BC The properties of detailed points, especially the right triangle with 30 degree angle
- 15. In the isosceles right triangle ABC, AC = BC, ∠ C = 90 degrees, ad bisects ∠ BAC, De is perpendicular to the point e AB = 15 cm, and the circumference of the triangle DBE is calculated In the isosceles right triangle ABC, the angle c is 90 degrees, AC = BC, ad bisection angle BAC, de vertical AB, perpendicular foot e, and ab = 15 cm?
- 16. As shown in the figure, △ ABC is a right triangle, ∠ C = 90 °, CA = CB, ad bisect ∠ BAC, de ⊥ AB, if AB = 18cm, calculate the circumference of △ DBE
- 17. As shown in the figure, in the triangle ABC, the midlines BD and CE intersect at the points o, e and F are the midpoint of OB and OC respectively, indicating that the quadrilateral defg is a parallelogram
- 18. As shown in the figure, in the RT triangle ABC, the angle ACB is equal to 90 degrees, and ab is equal to 6. Take AC and BC as the diameters to make semicircles, and the areas are recorded as S1 and S2 respectively, then what is S1 plus S2 equal to
- 19. It is known as follows: in △ ABC, ∠ cab = 120 °, ab = 4, AC = 2, ad ⊥ BC, D are perpendicular feet. Find the length of AD
- 20. Known; a = 201x + 2011, B = 2010x + 2012, C = 2010x + 2013. Find a ^ 2 + B ^ 2 + C ^ 2-ab-bc-ca