In the right triangle ABC, angle c is right angle, angle B ∶ angle c = 1 ∶ 4, find the degree of angle A The middle equation is still missing

In the right triangle ABC, angle c is right angle, angle B ∶ angle c = 1 ∶ 4, find the degree of angle A The middle equation is still missing

∵∠C=90° ∠B∶∠C=1∶4 ∴∠B=22.5°
∵∠C=90° ∠B=22.5° ∴∠A=67.5°

In RT △ ABC, ∠ C = 90 °, AC = 3, BC = 4, then the correct formula in the following is a.sinb = 3 / 5 b.cosb = 3 / 5 c.tanb = 3 / 5 d.tana = 3
Because we didn't learn it
So just ask for the answer

a

In RT △ ABC, ∠ C = 90 °, a = 31, C = 31, radical 2, solve this right triangle

The hypotenuse of an isosceles right triangle is twice the root of the right triangle
A = 31 is the right angle side, C = 31 √ 2 is the bevel side, C / a = √ 2
This is an isosceles right triangle with right side 31

As shown in the figure, in RT △ ABC, ∠ C = 90 °, D is a point on the edge of AC, de ⊥ AB is in E, de: AE = 1:2. Find SINB, CoSb, tanb

∵∵∵ a = ∵ a, ∵ AED = ∵ ACB, ∵ ABC ∵ ade, ∵ BC: AC = de: AE = 1:2, let BC = x, then AC = 2x, then AB = BC2 + ac2 = 5x, ∵ SINB = acab = 255, CoSb = bcab = 55x, tanb = ACBC = 2

Arithmetic sequence - 5, - 8, - 11 General term of

a1=-5,d=a2-a1=-3
So an = a1 + (n-1) d = - 3n-2
The general term is an = - 3n-2

Find the sum of the first 20 terms of the arithmetic sequence 8,5,2

The first term of arithmetic sequence is A1 = 8, and the tolerance is d = 5-8 = - 3
an=8+(n-1)×-3=-3n+11
∴a20=-3×20+11=-49
∴s20=(a1+a20)×20÷2
=(8-49)×20÷2
=-410

As shown in Figure 13, △ Abe and △ ACD are △ ABC along AB and AC respectively

As shown in the figure, △ Abe and △ ADC are formed by △ ABC folding 180 ° along AB and AC sides respectively. If ∠ 1: ∠ 2: ∠ 3 = 28:5:3, what is the degree of ∠ α? ∵ 1: ∠ 2: ∠ 3 = 28:5:3, ∵ set ∠ 1 = 28x, ∠ 2 = 5x, ∠ 3 = 3x, from ∠ 1 + ∠ 2 + ∠ 3 = 180

As shown in the figure, △ Abe and △ ACD are formed by △ ABC folding 180 ° along the sides of AB and AC respectively. If the degree of ∠θ is 50 °, then the degree of ∠ BAC is______ ．

The ∵ △ Abe is formed by △ ABC folding 180 ° along the edge of AB, ∵ △ ACD is formed by ∵ ABC folding 180 ° along the edge of AC, ∵ △ ACD is formed by ∵ ACB = ∵ ACD, ∵ ACD = ∵ e, and ∵ ACD + ∵ CAE = ∵ e + ∵ θ, ∵ EAC = ? θ = 50 °, ? BAE + ∵ BAC = 360 ° - 50 ° = 310 ° and ∵ BAC = 155 °

As shown in the figure, the side length of equilateral △ ABC is 6, ad is the middle line on BC side, M is the moving point on ad, e is a point on AC side. If AE = 2, the minimum value of EM + cm is ()
A. 27B. 4C. 37D. 1+27

Connect be and ad at the point g. ∵ △ ABC is an equilateral triangle, ad is the middle line on the edge of BC, ∵ ad ⊥ BC, ∵ ad is the vertical bisector of BC, ∵ the corresponding point of point c about AD is point B, ∵ be is the minimum value of EM + cm. ∵ point G is the calculated point, that is, point G coincides with point m, take the midpoint F of CE, and connect DF. ∵ equilateral ⊥ ABC's side length is 6, AE = 2, ∵ CE = ac-ae = 6-2 = 4, ∵ CF = EF = AE = 2, In the right angle △ BDM, BD = 12bc = 3, DM = 12ad = 332,... BM = BD2 + DM2 = 327,... Be = 43 × 327 = 27. The minimum value of ∵ EM + cm = be ∵ EM + cm is 27. So a

As shown in the figure, the side length of equilateral △ ABC is 6, ad is the middle line on BC side, M is the moving point on ad, e is a point on AC side. If AE = 2, the minimum value of EM + cm is ()
A. 27B. 4C. 37D. 1+27

Connect be and ad at the point g. ∵ △ ABC is an equilateral triangle, ad is the middle line on the edge of BC, ∵ ad ⊥ BC, ∵ ad is the vertical bisector of BC, ∵ the corresponding point of point c about AD is point B, ∵ be is the minimum value of EM + cm. ∵ point G is the calculated point, that is, point G coincides with point m, take the midpoint F of CE, and connect DF. ∵ equilateral ⊥ ABC's side length is 6, AE = 2, ∵ CE = ac-ae = 6-2 = 4, ∵ CF = EF = AE = 2, In the right angle △ BDM, BD = 12bc = 3, DM = 12ad = 332,... BM = BD2 + DM2 = 327,... Be = 43 × 327 = 27. The minimum value of ∵ EM + cm = be ∵ EM + cm is 27. So a