In the acute angle △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If C = 2A, then C The value range of a is () A. ( 2, 3) B. (1, 3) C. ( 2,2) D. (1,2)

In the acute angle △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If C = 2A, then C The value range of a is () A. ( 2, 3) B. (1, 3) C. ( 2,2) D. (1,2)

By sine theorem a
sinA=c
sinC,
∵C=2A∴c
sin2A=a
sinA=c
2sinAcosA,
∴c
a=2cosA,
When C is the largest angle, C ﹤ 90 ° a ﹤ 45 °
When B is the largest angle, B ﹤ 90 ° a ﹥ 30 °
∴30°<A<45°,
2cos45°<2cosA<2cos30°,
∴c
a∈(
2,
3)
Therefore, a

In the acute triangle ABC, the opposite sides of angle ABC are ABC respectively. If a = 1 and B = 2A, then the value range of B is

b/sin2A=a/sinA
b/a=sin2A/sinA=2cosA
Zero

Given that the symmetric point of point P (a + 1,2a-1) about the x-axis is in the first quadrant, find the value range of A In the angle ABC, if the length of three sides is 3, 1-2a, then the value range of a is Given that the ratio of the sum of the inner and outer angles of a polygon is 9: is this polygon? Polygon For a polygon, the sum of the other inner angles is 2750 ° except for the inner angles Given that the circumference of the angle ABC is 27, a, B, C are three sides of a 3-angle shape, and B + C = 2A, C = 1B in 2, calculate the value of ABC The circumference of the isosceles triangle is 21cm. The center line on one waist is used to calculate the length of each side of the isosceles triangle by two triangles with a circumference difference of 3cm There are three villages a, B and C. village B is to the west of Village C, village a is 20 ° north by east of village B, and 45 ° north by west of North Village of village C. what is the perspective of looking at village B and Village C from village a? In the angle a, B, C, ∠ A: ∠ ACB = 4:5: BD, CE are AC, AB, the height of the edge, BD, CE intersect with F, and calculate the degree of ∠ BFC The exterior bisector CD of the angle ABC and the extension of Ba intersect with the point D. try to prove that ∠ BAC is greater than ∠ B

The symmetry point of P is p '(a + 1, - 2A + 1)
Because in the first quadrant, then there is
a+1>0
-2a+1>0
So the value range of a is - 1
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Given the point m (1-A, 2A + 2), if the symmetric point of point m about the y-axis is in the third quadrant, then the value range of a is?

If M is symmetric about the Y axis in the third quadrant, then M itself is in the fourth quadrant
Therefore, the abscissa is greater than 0 and the ordinate is less than 0
1-a>0,a<1
2a+2<0,a<-1
So the solution set of inequality system is: a < - 1

Given the point m [1 - A, 2A + 2], if the symmetric point of point m about the x-axis is in the third quadrant, then the value range of a is____ If point m is on the bisector of the second and fourth quadrants, then a=____ ?

M [1 - A, 2A + 2], if the symmetric point of point m about the x-axis is in the third quadrant,
1-a<0
2a+2>0
a>1,a>-1
The value range of a is: a > 1
If point m is on the bisector of the second and fourth quadrants,
1-a=2a+2
3a=-1
a=-1/3

If the symmetric point of point m (1,2a-3) about the x-axis is in the first quadrant, then the value range of a is

The symmetric point of point m (1,2a-3) about the x-axis is (1,3-2a)
3-2a>0;
So a < 3 / 2;
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Mathematics of grade 2: if the symmetry point of P (a + 1,2a-1) about X axis is in the first quadrant, then the value range of a is given

P is in the fourth quadrant
So a + 1 > 0, 2a-1

If the intersection of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis, then the value range of a is______ .

∵ the function y = (A-2) x + 2a-3 is a linear function,
A-2 ≠ 0, that is, a ≠ 2
And ∵ the intersection point of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis,
∴2a-3>0,
A > 3
2,
The value range of a is: a > 3
2 and a ≠ 2
So the answer is: a > 3
2 and a ≠ 2

If the intersection of the image of the first order function y = (2a-3) x + 2-A and the Y axis is above the x-axis, and Y decreases with the increase of X, then the value range of a is?

Because the intersection of the image and the y-axis is above the x-axis
So 2-A > 0
A

If the intersection of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis, then the value range of a is______ .

∵ the function y = (A-2) x + 2a-3 is a linear function,
A-2 ≠ 0, that is, a ≠ 2
And ∵ the intersection point of the image of the first order function y = (A-2) x + 2a-3 and the y-axis is above the x-axis,
∴2a-3>0,
A > 3
2,
The value range of a is: a > 3
2 and a ≠ 2
So the answer is: a > 3
2 and a ≠ 2