In the acute triangle ABC, the edges of the three inner angles a, B and C are a, B, C respectively. Let m = (COSA, Sina), n = (COSA, - Sina), A = radical 7, and m * n = - 1 / 2 1) If B = 3, find the area of triangle ABC 2) Finding the maximum value of B + C

In the acute triangle ABC, the edges of the three inner angles a, B and C are a, B, C respectively. Let m = (COSA, Sina), n = (COSA, - Sina), A = radical 7, and m * n = - 1 / 2 1) If B = 3, find the area of triangle ABC 2) Finding the maximum value of B + C

M * n = cos2a = - 1 / 2, and the angle a is an acute angle, a = 60 ° is obtained. From the cosine theorem (including angle a), C = 1 (round off) (if the angle c is obtuse angle) or C = 2, area = 3 root sign 3 / 22) is known from the sine theorem B = 2 * (7) ^ (1 / 2) / (3) ^ (1 / 2) * SINB, C = 2 * (7) ^ (1 / 2) / (3) ^ (1 / 2) * sincb + C = 2 * (7) ^ (1 / 2) / (3)

It is known that ABC is the three inner angles of the acute triangle ABC, and the vector a = (Tana, - Sina) B = (1 / 2sin2a, CoSb) the angle between vector a and B is α (1) Verification: 0 ≤ α < π / 2 (2) Find the maximum value of the function f (α) = 2Sin ^ 2 (π / 4 + α) - radical 3 Cos2 α

（1）a=(tanA,-sinA),b=(1/2sin2A,cosB）
a●b=tanA*1/2sin2A-sinAcosB
=sinA/cosA*sinAcosA-sinAcosB
=sin A-sinAcosB
=sinA(1-cosB)
∴sinA(1-cosB)>0
Tana * CoSb + Sina * 1 / 2sin2a
=sinAcosB/cosA+sin AcosA
∵ a, B, C are the three inner angles of an acute triangle
sinAcosB/cosA+sin AcosA>0
The vectors a and B are not collinear
The angle range of vector a and B is (0, π / 2)
(2) F (x) = 2Sin ^ 2 (π / 4 + x) - radical 3 cos2x
=1-cos (π / 2 + 2x) - radical 3 cos2x
=Sin2x radical triple 2x + 1
=2sin（2x-π/3）+1
∵ x∈[0,π/2）
∴2x-π/3∈[-π/3,2π/3）
The maximum value of F (x) is 3
So use x instead. I'm so tired

If a, B, C are the three internal angles of the acute angle △ ABC, vector P=（1+sinA，1+cosA）， Q = (1 + SINB, - 1-cosb) P and The angle of Q is () A. Acute angle B. Obtuse angle C. Right angle D. Not sure

In the acute angle △ ABC, sina ＞ CoSb ＞ 0, SINB ＞ Cosa ＞ 0,
So there are
p•
q=（1+sinA）（1+sinB）-（1+cosA）（1+cosB）＞0，
At the same time, it is easy to know
P and
Q direction is not the same, so
P and
The angle between Q is acute
Therefore, a

If a, B, C are the three internal angles of the acute angle △ ABC, vector P=（1+sinA，1+cosA）， Q = (1 + SINB, - 1-cosb) P and The angle of Q is () A. Acute angle B. Obtuse angle C. Right angle D. Not sure

In the acute angle △ ABC, sina ＞ CoSb ＞ 0, SINB ＞ Cosa ＞ 0,
So there are
p•
q=（1+sinA）（1+sinB）-（1+cosA）（1+cosB）＞0，
At the same time, it is easy to know
P and
Q direction is not the same, so
P and
The angle between Q is acute
Therefore, a

It is known that the triangle ABC is an acute triangle, and the three inner angles are a, B, C. the known vector p = (2-2sina, cosa + Sina) q = (1)+ sinA.cosA -If P is perpendicular to Q, find the size of the inner angle A

∵ vector p ⊥ vector Q, ᙽ (2-2sina) * (1 + Sina) + (COSA + Sina) * (cosasina) = 0
2*1+2sinA-2sinA-2sin^2A+cos^2A-sin^2A=0.
1+1-2sin^2A+cos2A=0.
1+cos2A+cos2A=0.
2cos2A=-1.
cos2A=-1/2.
2A=120°
∴∠A=60°.

Given that the angles a, B, C are the three inner angles of △ ABC, the opposite sides of which are a, B and C respectively, if the vector M = (- cosa / 2, Sina / 2), the vector n = (COSA / 2, Sina / 2), A = 2 √ 3, and vector m * vector n = 1 / 2 1, if the area of △ ABC s = √ 3, find the value of B = C 2. Find the value range of B + C

1 vector m · vector n = 1 / 2
(-cosA/2,sinA/2)*(cosA/2,sinA/2)=1/2
=-(cos²A/2-sin²A/2）
=-cosA
cosA=-1/2
A=120°
1 / 2 * b * c * sin120 = √ 3, so B = C = 2
Sina = radical 3 / 2
A / Sina = B + C / SINB + sinc = 2R = 8 root sign 3
sinb=sin(60-c)
SINB + sinc = radical 3 / 2cosc + 1 / 2sinc
B + C = 8 root 3 (root 3 / 2cosc + 1 / 2sinc) = 8 root sign 3sin (c + 60) C belongs to (0,60)
Twelve

A is an inner angle of the triangle ABC. If Sina + cosa = 1 / 5, then the shape of the triangle

(Sina + COSA) ^ 2 = 1 + 2sina * cosa = 1 / 25, and the simultaneous equations of sina + cosa = 1 / 5 give - 3 / 5 and 4 / 5. Since the sine value of triangle is Sina > 0, cosa = - 3 / 5 is an obtuse triangle

It is known that the acute angle a is an inner angle of the triangle ABC, and a, B, C are the corresponding sides of each internal angle in the triangle, if (Sina) ^ 2 - (COSA) ^ 2 = 1 / 2 Then: A b+c=2a B b+c

(sinA)^2-(cosA)^2=1/2
(sinA)^2-[1-(sinA)^2]=1/2
2(sinA)^2=3/2
sinA=√3/2
cosA=1/2
∠A=60°
∠B+∠C=120°
1: When ABC is an equilateral triangle, a = b = C, B + C = 2A
2: When ABC is a right triangle, a ^ 2 + B ^ 2 = C ^ 2
2b=c
2b+b=b+c>a
a^2=c^2-b^2=(c+b)(c-b)=3b*b=3b^2
a=√3b
2a=2√3b
b+c=3b
2a>b+c
In both cases, the answer should be C: B + C less than or equal to 2A

A is an inner angle of the triangle ABC if Sina + cosa = 12 25, then the shape of the triangle is () A. Acute triangle B. Obtuse triangle C. Isosceles right triangle D. Isosceles triangle

∵ Sina + cosa = 1225, ᙽ the square of both sides leads to (Sina + COSA) 2 = 144625, that is, sin2a + 2sinacosa + cos2a = 144625, ∵ sin2a + cos2a = 1,

As shown in the figure, there is a circle O and two regular hexagons T1, T2. The six vertices of T1 are on the circumference, and the six edges of T2 are tangent to circle O (we call T1 and T2 respectively the inscribed regular hexagon and the circumscribed regular hexagon of circle O) (1) Let the side lengths of T1 and T2 be a and B respectively, and the radius of circle o be r, and then calculate the values of R: A and R: B; (2) Find the value of the area ratio S1: S2 of the hexagon T1 and T2

(1) Six congruent equilateral triangles can be obtained by connecting the six vertices of the center O and T1
So r: a = 1:1;
Connect the two adjacent vertices of the center O and T2 to form an equilateral triangle with a high radius of circle o,
So r: B = Ao: Bo = sin 60 degrees=
3：2；
(2) The side length ratio of T1: T2 is
3: So S1: S2 = (A: b) 2 = 3:4