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In the triangle ABC, ∠ C = 90 degrees, CD ⊥ AB, ∠ B = 56 degrees, then ∠ DCA
Because ∠ C = 90 degrees, CD ⊥ AB, so ∠ ACD + ∠ BCD = 90 degrees. In triangular CBD, ∠ DBC + ∠ DCB = 90 degrees
By synthesizing the above two formulas, we can get ∠ DCA = 56 degrees
Given C = 68, a = 34 ° and B = 56 ° find the area of a sum triangle ABC
Because the angle c = 180-34-56 = 90 degrees
So a = 68sin34 °≈ 38
b=68sin56°≈56.4
S triangle ABC = 38 * 56.4 / 2 = 1071.6
It is known that the equilateral △ ABC, P is on the ray ba. Baap = n, (n ≠ 1) (1) as shown in Fig. 1, when n = 2, passing point P, PF ⊥ BC is on F and intersecting point AC is on point E. verification: AE = EC; (2) as shown in Fig. 2, point D is on the extension line of BC, BC = CD, PC = PD, and the value of N; (3) if point P is on the ray Ba, D is on the straight line BC, PC = PD, then ACCD = PD___ (expressed by a formula containing n)
(1) It is proved that ∵ △ ABC is an equilateral triangle, ∵ B = ∵ BAC = ∵ C = 60 °, ∵ PF ⊥ BC, ∵ P = 30 °, ∵ AEP = ∵ BAC - ∵ P = 30 °, P = ∵ AEP, ∵ AP = AE, ∵ n = 2, ∵ ab = 2AP, and ab = AC, ∵ AC = 2ae, ∵ AE = EC. (2) it is proved that as shown in Fig. 2, the extension line of PM ∵ AC intersection BC through P is an equilateral triangle, ? B = ∵ BAC = ∵ ACB = 60 °, M= In △ PBC and △ PMD, ∠ B = ∠ M = 60 °; (AAS), ∥ BC = DM, ∥ BC = CD, ∥ BC = CD = DM = 13bm, and ∥ BC = Ba, BM = BP, ∥ BP = 3ba, ∥ AP = 2Ab, ∥ n = baap = 12; (3) as shown in Figure 3, BC = DM is obtained by the same method as (2), so n = baap = ACCD + DM = ACCD + AC, ∥ ACCD = n1-n.so the answer is: n1-n
Given the parametric equation of a curve, how to find the surface equation of its rotation around the X axis?
The parametric equation is x = 2, y = 2T, z = 3T. What is the surface equation rotating around the x-axis? Please tell me the related steps and formulas
Given that the parameter equation is very special, it is located in the x = 2 plane, so the result of rotation is a group of concentric rings (the range of variation with the parameter t is a ring or disk or the whole x = 2 plane), and the equation of concentric circles is y ^ 2 + Z ^ 2 = 13T ^ 2; if x is also a linear function of T, the result of rotation is a circular mesa or cone (or opposite cone)
Four mixed fractional operations,
（1）3/4×4/9÷（4/5－3/10）
（2）1/4+3×7/6
（3）3/7×4/5+4/7×2/5
（4）5/7÷3－2/21
（1）3/4×4/9÷（4/5－3/10）
=1/3÷(8/10-3/10)
=1/3÷5/10
=1/3÷1/2
=2/3
（2）1/4+3×7/6
=1/4+7/2
=1/4+14/4
=15/4
（3）3/7×4/5+4/7×2/5
=4/5*(3/7+2/7)
=4/5*5/7
=4/7
（4）5/7÷3－2/21
=5/21-2/21
=3/21
=1/7
Let f (x) = LG [2 / (1-x) + a] defined on (- 1,1) be an odd function, then f (x) 1? Then there is no solution?
∵ f (x) is an odd function ∵ f (x) + F (- x) = 0 ∵ LG [2 / (1-x) + a] + LG [2 / (1 + x) + a] = 0, and (a + 1) [4 / (1-x & sup2;) + A-1] = 0 holds ∵ a + 1 = 0, a = - 1 ∵ f (x) = LG (1 + x) / (1-x), X ∈ (- 1,1) Let f (x) < 0, then (1 + x) / (1-x) < 1, and X ∈ (- 1,0) chooses a
If x and y are real numbers and 1 / 2 ≤ x ^ 2 + 4Y ^ 2 ≤ 2, find the maximum and minimum of x ^ 2-2xy + 4Y ^ 2
Let x = rcost, y = (R / 2) Sint. [1 / (√ 2) ≤ R ≤ √ 2.]. Then z = x ^ 2-2xy + 4Y ^ 2 = R ^ 2 * [2 - (sin2t)] / 2. Obviously, (z) min = 1 / 4, (z) max = 3
Solving ten problems of binary linear equations
Do not apply the problem, is to solve the equation, a little bit difficult, like with brackets and fractions that kind of
1.36x+77y=7619 (47x-1/2y)+5=799 2.13x-42y-3=-2717 42x+3/5y-3=1333 3.28x+28y=3332 52x-y+2=4628 4.62x-2+98y=-2564 46x-5+y=2024 5.79x-76y=-4388 26x-5/9y+28=832 6.63x-40y=-821 42x-y=546 7.69x-96y=-1209 42...
How to find the derivative of a complex function? For example, the derivative of X + Xi
The general form of the function of complex variable is f (z) = u (x, y) + IV (x, y). If the function of complex variable is differentiable, it needs to satisfy the Cauchy Riemann equation, that is, u'x = v'y, u'y = - v'x. in your example, u = v = x, then u'x = 1, V'y = 0, and u'x ≠ v'y, so it is not differentiable. If the function of complex variable is differentiable, then its derivative f '(z) = u'x + iv'x
Given a, B, a + B isochromatic, a, B, AB isocratic, and 0
a. B, a + B isochromatic
2b=a+a+b=2a+b
b=2a
a. B, AB, etc
b^2=a*ab=a^2b
Equal ratio cannot have 0
So B is not equal to 0
So B = a ^ 2
b=2a
A is not equal to 0
So a = 2, B = 4
So 0
RELATED INFORMATIONS
- 1. There is a point P in the equilateral triangle ABC. The distances from point P to vertex a, B and C are 3, 4 and 5 respectively. The degree of angle APB is calculated
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