As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
Let ∠ Abe = 2x ° and get 2x + 21 = 5x-21. The solution is x = 14 and ﹥ ABC = 14 °× 7 = 98 °. The degree of ﹥ ABC is 98 degrees. So the answer is 98 degrees
RELATED INFORMATIONS
- 1. As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
- 2. Given that the point C is any point on the line AB (C does not coincide with a or b), AC and BC are taken as sides respectively, and the equilateral triangle ACD and equilateral △ BCE, AE and C are made on the same side of ab D intersects with m, BD intersects with CE with n The results show that: (1) ace ≌ DCB; (2) ACM ≌ DCN; (3) Mn ∥ ab
- 3. C is a point on the line AB, with AC and BC as edges, make equilateral △ ACD and equilateral △ BCE on the same side of AB, AE and CD intersect at m, BD and CE intersect at n It is proved that (1) △ MCN is an equilateral triangle. (2) if AC: CB = 2:1, then de ⊥ CE
- 4. As shown in the figure, the line AB = 4, C is a moving point on the line AB, with AC and BC as sides, the equilateral triangle ACD and equilateral triangle BCE are made, Finding the minimum value of de
- 5. Point C is an arbitrary point on the line AB, with AC and BC as sides respectively as regular triangles. Note that the area of △ BCE is S1, and the area of △ ACD is S2 If C is the golden section of AB and AC > BC, S1: S2 is obtained
- 6. As shown in the figure, ab ∥ CD, ∠ ACB = 90 °, e is the midpoint of AB, CE = CD, de and AC intersect at F.; De is perpendicular to AC, and angle ACD = angle As shown in the figure, it is known that AB / / CD, ∠ ACB = 90 °, e is the midpoint of AB, CE = CD, de and AC intersect at F, and the verification is as follows:
- 7. When the bulb is always marked with "220 V 40 W", the current passing through the filament is - A, one degree of electric energy for the lamp to work normally for - hours There must be a way to solve the problem!
- 8. AC power problem The maximum value of alternating current (sine cosine type) is 250 root 2 v. if a fuse is connected, the maximum current allowed by the fuse is 10a, and the maximum output power allowed is 5000W. What's wrong? The voltage is 250 bar No. 2, and the fuse is the effective value, so the maximum current is 10 bar No. 2, and the maximum power is 250 bar No. 2 * 10 bar No. 2 = 5000?
- 9. What is a ton represented by a symbol
- 10. Conversion of capacitance units? Capacitance units include h, uh, pH, etc. What's the conversion between them? It seems that there is any relationship between MH and NH?
- 11. As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
- 12. As shown in the figure, make equilateral triangles △ abd, △ EBC, △ fac on the same side of BC with the three sides of △ ABC respectively, and prove that the quadrilateral ADEF is a parallelogram
- 13. As shown in the figure, D is a point on side AC of triangle ABC, CD = 2ad, AE is perpendicular to BC and E, if BD = 8, Tan angle abd = 3 / 4, find the length of AC
- 14. In triangle ABC, ad is perpendicular to BC and D, be is the middle line, and the angle EBC is 30 degrees
- 15. As shown in the figure, ad and be are the heights of BC and AC respectively in ABC
- 16. As shown in the figure △ ABC, ∠ BAC = 90 & ordm;, ad ⊥ BC, ∠ Abe = ∠ EBC, EF ⊥ BC, FM ⊥ AC, DF = FM
- 17. As shown in the figure △ ABC, ad ⊥ BC, be ⊥ AC, BF = AC, if ∠ EBC = 25 °, then ∠ ACF=______ .
- 18. As shown in the figure, D is a point inside the triangle ABC, connecting BD and ad, taking BC as the edge, making an angle outside the triangle ABC, CBE = angle abd, BCE = angle bad, be, CE intersecting at point E, connecting de. prove that the triangle DBE is similar to the triangle ABC
- 19. It is known that, as shown in the figure, D is a point inside △ ABC connecting BD and ad, with BC as the edge, outside △ ABC, make ﹥ CBE = ﹥ abd, ﹥ BCE = ﹥ bad, be and CE cross to e, connecting de. (1) verification: bcab = bebd; (2) verification: ﹥ DBE ﹥ ABC
- 20. As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reason