In △ ABC, a = 12, B = 13, C = 60 ° the solution of this triangle is () A. No solution B. one solution C. two solutions D. indeterminate

According to the cosine theorem, C2 = A2 + b2-2abcosc, because a = 12, B = 13, C = 60 °, C = 157

Ask for a math problem of senior two
In △ ABC, a = 3, B = radical 3, a = 30 ° are known to solve triangles

Sine theorem: A / Sina = B / SINB = C / sinc
That is: 3 / sin30 = radical, 3 / SINB = C / sinc
So: SINB = root 3 * sin30 / 3 = root 3 / 6
Angle B is not a special angle
The cosine theorem is a ^ 2 = B ^ 2 + C ^ 2-2bc cosa
9 = 3 + C ^ 2-2 radical 3 C * radical 3 / 2
c^2-3c-6=0
C = (3 + radical 33) / 2

Solve mathematical problems, high two solution triangle, seek master solution! Thank you very much!
In △ ABC, D is the point on BC, BD = 1 / 2dc, ∠ ADB = 120, ad = 2. If the area of △ ADC is 3 - root 3, calculate ∠ BAC
Solved, 60 degrees

Difficult

Given the function f (x) = 2cosxsinx - 2cos2x + 1, X is a real number
Q:
Monotone increasing interval of function f (x);
Find the maximum and minimum values of the function in the interval of 8 / π to 3 / 4 π

F (x) = 2cosxsinx-2cos2x + 1 = sin2x - (2cos2x-1) = sin2x cos2x = root 2Sin (2x-45) 360n-90 ≤ 2x-45 ≤ 360n + 90, 180n-22.5 ≤ x ≤ 180n + 67.5, n ∈ integer 2.22.5 ≤ x ≤ 1350 ≤ 2x-45 ≤ 225, so in 2x-45 = 90, that is, x = 67

1. Given the set a = {x | - 2 ≤ x ≤ 5}, B = {x | - M + 1 ≤ x ≤ 2m-1} and a contains B, find the value range of real number M
2. Let a = {- 1,1}, B = {x | X & sup2; - 2aX + B = 0}, if B ≠ an empty set and B is contained in a, find the value of a and B

1. From the known solution: - 2 ≤ m + 1 and 2m-1 ≤ 5, we get: - 3 ≤ m ≤ 22. From the known solution: B = {1} or {- 1} or {- 1,1} we also know: B ≠ empty set, so: B ≤ A & sup2; when B = A & sup2; - 2aX + B = 0, we get: a = 1 or a = - 1; b = 1. When b > A, we get x = 1 and x = - 1 into X & sup2; - 2aX +

1, set a = {x | x ^ 2-3x + 2 = 0}, B = {x | x ^ 2-ax + A + 1 = 0}, C = {x | x ^ 2-mx + 2 = 0}, if a ∪ B = a, a ∩ C = C, find the value of a, M
2. Given the function y = x ^ 2 + ax + B, a = {x | x ^ 2 + ax + B = 2x} = {2}, find the value of a, B and the explanation of quadratic function y
3, known set a = {x | 4 ≤ - 2x ≤ 8}, set B = {x | x-a ≥ 0}
(1) If a is a subset of B, find the range of A;
(2) If the complete set u = R and a is a subset of the complement of B, the range of a is obtained
4. Let s be a set which satisfies the following two conditions: (1) 1 &; s; (2) if a ∈ s, then 1 / 1-A ∈ s
It is proved that: (1) if a ∈ s, then 1-1 / a ∈ s
(2) If 2 ∈ s, then there must be two other numbers in s, and write these two numbers
Please write down the process of solving the problem. I can understand it in class, but I don't know why I can't do it

1: A = {x | x ^ 2-3x + 2 = 0}, a = {1,2} AUB = a, substituting x = 1 into x ^ 2-ax + A + 1 = 0, a has no solution, substituting x = 2 into x ^ 2-ax + A + 1 = 0, a = 3A ∩ C = C, substituting x = 2 into x ^ 2-mx + 2 = 0, M = 32: the root of x ^ 2 + ax + B = 2x is 2, which is the multiple root of x ^ 2 + (A-2) x + B = 0 (X-2) ^ 2 = x ^ 2-4x + 4, the contrast coefficient is: - 4 = A-24 =

Let f (x) = x & sup2; Cos θ + 2Sin θ - 1, θ∈ (0, π). If f (x) is an increasing function in the interval [- 1, √ 3], find the value range of θ
If the function f (x) satisfies f (x) + F (2a-x) = 2B for any X in the domain of definition, then the image of function y = f (x) is symmetric with respect to point (a, b)
(1) Given that the image of the function f (x) = x & sup2; + MX + m / X is symmetric with respect to (0,1), the value of real number m is obtained
(2) Given that the image of function g (x) on (- ∞, 0) ∪ (0, + ∞) is symmetric with respect to point (0,1), and when x ∈ (0, + ∞), G (x) = x & sup2; + ax + 1, find the analytic expression of function g (x) on X ∈ (- ∞, 0)
(3) Under the conditions of (1) and (2), if G (x) < f (T) is constant for real numbers x < 0 and T > 0, the value range of real number a is obtained
3 if f (x) = ax + 1 / - x + 2 is an increasing function in the interval (- 2, + ∞), then the value range of a
If the equation of the symmetry axis of the image with the function f (x) = log3 | 2x + a | is x = 2, then the constant a=

According to this formula, we can discuss (1) because it is an increasing function from the interval - 1 to the root 3, the axis of symmetry of the function must be on the left side of - 1 (from the image), and because the axis of symmetry = - B / 2a, when θ is from 0 to π / 2

In triangle ABC, if a > b > C, a = 2c, B = 4, a + C = 8, then the lengths of a and C are?
First floor. First of all, let's not talk about high school and primary school
If a = 2C is the length, can it be true?
It's ridiculous

Because a + B + C = 180 ° a = 2C so 3C + B = 180 ° so B = 180 ° - 3C because 4 / SINB = A / Sina = C / sinc so a = 4sina / SINB C = 4sinc / SINB because a + C = 8 so (Sina + sinc) / SINB = 2 so sin2c + sinc = 2Sin (180 ° - 3C) so 2sinccosc + sinc = 2 (sinccos2c + cossin2c

7x-5y-23≤0
x+7y-11≤0
4x+y+10≥0
(1) Finding the maximum value of Z = 4x + 3Y
(2) Finding the maximum value of Z = 4x-3y
(3) Finding the maximum value of Z = x ^ 2 + y ^ 2

Provide your own ideas
Take (1) as an example
Let z = a (7x-5y) + B (x + 7Y)
There are 7a + B = 4
7b-5a=3
Find out a, B, use inequality knowledge to solve
Another way of thinking:
Draw a picture, draw the range of X and Y (surrounded by three lines), then draw the objective function, and then calculate the intersection point and vertex value

Ask about the mathematics problem of solving triangle
In triangle ABC, if a ^ 2 / b ^ 2 = Tana / tanb, what is the shape of triangle ABC?

Because a ^ 2 / b ^ 2 = Tana / tanb
So a ^ 2 / b ^ 2 = (Sina / COSA) / (SINB / CoSb)
So a ^ 2 / b ^ 2 = sinacosb / sinbcosa
So a ^ 2sinbcosa = B ^ 2sinacosb
According to the sine theorem, divide sina on the right by sina on the left and SINB on the left by SINB on the right
2RacosA=2RbcosB
So acosa = bcosb
So 2rsinacosa = 2rsinbcosb
So sin2a = sin2b
So 2A = 2B or 2A + 2B = 180 degrees
So a = B or a + B = 90 degrees
So the triangle is an isosceles triangle or a right triangle

In △ ABC, a = 12, B = 13, C = 60 ° the solution of this triangle is () A. No solution B. one solution C. two solutions D. indeterminate