The three inner angles of the acute triangle ABC are known A, B, C

Edges a and b are the two roots of the equation x 2-2( 3) x + 2 = 0, so we know that a = 3 + 1, b = 3-1 or b = 3 + 1, a = 3-1

The angle a,b satisfies the relation 2sin(a+b)-(3)=0, then.

Since sin(a+b)=sinc, then sinc = 3 2, then c = 60° or 120°

1. When c=60°, it is obtained by the cosine theorem, c 2=a 2+b 2-2abcosc=4+2 3+4-2 3-2*2*1 2=6

So c= 6

2. When c = 120°, it is obtained by the cosine theorem, c 2 = a 2 + b 2-2abcosc = 4 + 2 3 + 4-2 3 + 2 * 2 * 1 2 = 10

So c= 10

3、s（abc）=1/2absinc=1/2*2*√3/2=√3/2

Then a+b=2 3, ab=2, the solution gives a= 3-1, b= 3+1sin(a+b)=sinc= 3 2, and the solution gives c=60 degrees.

c 2 = a 2 + b 2-2ab * cosc = 8-4 2 = 6, the solution is c = 6

sabc=ab*sinc/2=√3/2

Let the opposite sides of the acute triangle ABC be and A=2BSINAFind the size of b. Obtain the value range of cosa+sinc.

Solution 1: Find the size of b.

The perpendicular CD of AB is made through point C, where CD crosses AB at point D.

In the triangular ACD, cd=acsincad=bsina

In the triangular BCD, BC=A=2BSINA

So sinb=cd bc=bsina 2bsina=1 2

Because the triangle ABC is an acute triangle, the angle b is an acute angle, and sinb = 1 2 and b = 30°

Solution 2: Find the value of COSA+SICC Fan Yin's hand circumference.

Since b = 30° and abc is an acute triangle, a+c = 180°-30° = 150°, this known condition is used to get the answer.

cosa=sin(90°-a)

sicc=sin(180°-b-a)

cosa+sicc=sin(90°-a)+sin(180°-b-a)

Using the trigonometric relationship: sin sin 2sin[( 2]·cos[( 2]).

cosa+sicc=sin(90°-a)+sin(180°-b-a)

2sin]·cos

2sin(120°-a)·cos(-30°)

3sin(120°-a)……The first ......... of the letter fight1) formula.

Because a is an acute angle, i.e. 0 a 90°

So 30° 120°-a 120°

So 1 slip 2 sin(120°-a) 1

So 3 2 3sin(120°-a) 3

Therefore, the value range of COSA+SICC is (3, 2, 3).

cosa sinb, cosa cos[(2)-b], a,b are both acute angles, a,( 2)-b (0, 2).

From the function y=cosx is a subtraction function on (0, 2), a<(2)-b, i.e., a+b< 2, from the sum theorem of the inner angles of the triangle, it can be seen that c> 2, abc is an obtuse triangle with c as the obtuse angle.

cosa>sinb

sin(90-a)>sinb

Because a, b are acute angles, and 90-a is also acute angles.

So 90-a>b

a+b<90

So c=180-(a+b)>90

It is an obtuse triangle.

cosa sinb, cosa cos[(2)-b], a,b are both acute angles, a,( 2)-b (0, 2).

From the function y=cosx is a subtraction function on (0, 2), a<(2)-b, i.e., a+b< 2, from the sum theorem of the inner angles of the triangle, it can be seen that c> 2, abc is an obtuse triangle with c as the obtuse angle.

Solution: cosa sinb, and a and b are both acute angles, a b 45°

c＞90°.

ABC is an obtuse triangle.

cos²a+sin²a=1

The solution yields cos a = 1 4, sin a = 3 4, sina = 3 2 due to abc is.

Acute triangle.

So a= 3

s=1/2*bcsina=√3/4*bc

Composed. Cosine theorem.

b²+c²-a²=2bccosa

i.e. b + c = 7 + bc

And because of b + c 2bc

So 7+bc 2bc

BC 7 Area s 7 3 4

The maximum value is 7 3 4

Answer: (1).

cos(a-c)-cos(a+c)=3/22sinasinc=3/2

sinasinc=3/4

Because: b 2 = ac, so: a b = b c

According to the sinusoidal theorem: a sina = b sinb sinb so: a b = sina sinb sinb (sinb) 2=sinasinc=3 4sinb = 3 2

b=60° or b=120° (do not match the acute triangle and need to be rounded) because the triangle ABC is an acute triangle, so b=60°(2)sinasinc

sinasin(180°-60°-a）

sina (sin120 ° cosa-cos120 ° sina) = 3sinacosa 2 + sinasina 2 = 3 4 so: 3sin2a-cos2a = 2

Combined: (sin2a) 2+(cos2a) 2=1 to get the solution:

sin2a=√3/2，cos2a=-1/22a=120°,a=60°

Combine (1) to know: a=b=c=60°

So a=b=c=3

The circumference is 3 3

According to the summation formula sn=n(a1+an) 2, we can know that 3(a+c) 2=180°, that is, a+c=120°, i.e., b=60°

According to the problem abc is an acute triangle, it can be seen that the value range of a is (30°, 90°), and the value range of c is (30°, 90°).

When a infinity approaches 30°, c infinitely approaches 90°, there can be a right triangle that approaches a=2, b=2 3, c=4, and the triangle has an area of 2 3, but cannot be reached, so the option cd is excluded.

When a= b= c=60°, the triangle is an equilateral triangle with sides of 2 3 and an area of 3 3, which meets the requirements of the question.

In summary, the value range of the triangle area is b, (2 3, 3 3].

b, abc is an equal difference series, so b is 60 degrees. The largest area is an equilateral triangle, and the limit on the other side is a right triangle.