In the triangle ABC, the opposite sides of the inner angle ABC are a, B, C. If a ^ 2-B ^ 2 = root 3 * BC, sinc = 2 radical 3 * SINB, then a= As the title

In the triangle ABC, the opposite sides of the inner angle ABC are a, B, C. If a ^ 2-B ^ 2 = root 3 * BC, sinc = 2 radical 3 * SINB, then a= As the title

From cosine theorem cosa = (B? + C? - a?) / 2BC = [C? - (√ 3) BC] / 2BC = C / (2b) - (1 / 2) √ 3 (*) by substituting the sine theorem C / b = sinc / SINB = 2 √ 3 (*), cosa = (√ 3) / 2 ∵ 0 is obtained by substituting the cosine theorem cos a = (B ﹣ C ∵ 3) / 2BC = [C ∵ 3)

In the triangle ABC, the sides of ABC are ABC, B = 3 / π, cosa = 4 / 5, B = radical 3. Find the value of sinc and the area of triangle ABC

COSA=4/5 sinA=√(1-16/25)=3/5
√3/(√3/2)=a/(3/5) a=6/5
sinC=sin(180-A-B)
　　=sin(A+B)
　　=sinAcosB+cosAsinB
　　=3/5*1/2+4/5*√3/2
　　=(3+4√3)/10 ≈0.9928
The area of triangle ABC is s = 1 / 2absin C = 1 / 2 * 6 / 5 * √ 3 * (3 + 4 √ 3) / 10 ≈ 1.0318


In the triangle ABC, the opposite sides of a, B and C are a, B, C, B = 60 degrees, cosa = 4 / 5, B = root sign 3.1: find the value of sinc 2: find the triangle In the triangle ABC, the opposite sides of a, B and C are a, B, C, B = 60 degrees, cosa = 4 / 5, B = radical 3 1: Find the value of sinc 2: Find the area of triangle ABC

Because cosa is positive, then the angle a is an acute angle, then Sina = 3 / 5, Sina = sin (180-a-b) = sin (a + b) = sinacosb + sinbcosa, and then substitute the value into the value. The value of a can be obtained from the positive theorem, and then the answer can be obtained by S = absinc / 2

It is known that: A, B, C are the inner angles of the triangle ABC, a, B, C are their opposite sides respectively, vector M = (radical 3, cos (Wu - a) - 1), n = (c It is known that a, B, C are the inner angles of the triangle ABC, a, B, C are their opposite sides respectively, and the vectors M = (radical 3, cos (Wu - a) - 1), n = (COS (U / 2-A), 1), m, vertical n. (1) If the length of the root of a is 2, find the size of B

Vector M = (√ 3, - cosa-1), n = (Sina, 1)
1)m*n=√3sinA-cosA-1=0,
2sin(A-π/6)=1,
A-π/6=π/6,
A=π/3.
cosB∴B>A,sinB>sinA,
sinB=√6/3,
b=asinB/sinA=4√2/3.

In the case of ABC, where (cosin / 2) is the root of ABC, 2 is the inner edge of ABC (1) Find the size of angle c (2) If ABC is an equal sequence, find the value of sina

(1) It is known that √ 2Sin 2 (C / 2) + cos (C / 2) = √ 2, that is √ 2 [1-cos 2 (C / 2)] + cos (C / 2) = 0, ∵ C ≠ 180 °, cos (C / 2) ≠ 0. Divide cos (C / 2) into cos (C / 2) + 1 = 0, or cos (C / 2) = √ 2 / 2, ∵ C /

In the triangle ABC, the angles a, B and C are a, B, C and satisfy the root sign 5 / 5 of cosa / 2 = 2 times In the triangle ABC, the edges of angles a, B and C are a, B, C and satisfy the root sign 5 / 5 of cosa / 2 = 2 times, and vector AB times vector AC = 3 (1) Find the area of triangle ABC (2) If B + C = 6, find the value of A

Vector AB * vector AC = |||||||ac* cosa = BC * cosa = 3
Cosa = 2 * (COSA / 2) ^ 2-1 = 0.6, b * C = 5
sinA=0.8
Triangle area = 0.5 * b * c * Sina = 2
b+c=6
We get b = 5, C = 1 (or C = 5, B = 1)
The cosine theorem a ^ 2 = B ^ 2 + C ^ 2-2cosa * BC gives a = 2 √ 5

In a triangle, if cosa / 2 = 2, root sign 5 / 5, vector AB * vector AC = 3, calculate the area of ABC; if AC + AB = 6, find the value of BC

First question:
∵cos（A/2）＝2√5/5,∴［cos（A/2）］^2＝4/5,∴cosA＝2［cos（A/2）］^2－1＝3/5＞0,
A is an acute angle, ν Sina = √ [1 - (COSA) ^ 2] = √ (1-9 / 25) = 4 / 5
∵ cosa = vector ab · vector AC / (︱ vector ab ︱ vector AC ︱ cosa = 3 / 5,
ν vector ab · vector AC / (︱ vector ab ︱ vector AC ︱ 3 / 5,  ab × AC) = 3 / 5,  ab × AC = 5
The area of △ ABC = (1 / 2) ab × acsina = (1 / 2) × 5 × (4 / 5) = 2
Second question:
According to the cosine theorem, there are:
BC^2＝AB^2＋AC^2－2AB×ACcosA＝（AB＋AC）^2－2AB×AC－2AB×ACcosA
＝36－2×5－2×5×（3/5）＝26－6＝20.
∴BC＝2√5.

In △ ABC, the edges of angles a, B and C are a, B and C respectively, and cosa / 2 = 2, radical 5 / 2, AB vector · AC vector = 3 (1) Find the area of △ ABC? (2) if B + C = 6, find the value of a?

(1)
∵ cosa / 2 = 2 √ 5 / 5, [denominator is 5, so you gave 2, wrong]
∴cosA=2cos²A/2-1=8/5-1=3/5
sinA=√(1-cos²A)=4/5
∵ AB vector · AC vector = 3
∴|AB||AC|cosA=3 ,
∴3/5|AB||AC|=3,|AB||AC|=5
The area of △ ABC
S=1/2|AB||AC|sinA=1/2*5*4/5=2
(2)
From (1) we know that CB = 5 and B + C = 6
∴b²+c²=(b+c)²-2bc=36-10=26
According to the cosine theorem
∴a²=b²+c²-2bccosA=26-2*5*3/5=20
∴a=2√5

6. Given that a and B are the two inner angles of the triangle ABC, the vector a = {radical 2 * cos (a + b) / 2} I + {sin (a-b) / 2} J, where I and j are mutually perpendicular unit vectors. If the absolute value a = root sign 6 / 2 (1) ask whether Tana * tanb is a fixed value, ask for it, otherwise explain the reason, (2) find the maximum value of Tanc and judge the shape of triangle at this time

Absolute value a = radical 6 / 22cos 2 (a + b) / 2 + sin 2 (a-b) / 2 = 3 / 21 + cos (a + b) + [1-cos (a-b)] / 2 = 3 / 2cosacosb sinasinb - [cosacosb + sinasinb] / 2 = 0cosacosb / 2 = 3sinasinb / 2tantanb = 1 / 3tanc = - Tan (a + b) = - [Tana + tanb] / (1-tanatanb

It is known that a and B are the two inner angles of the triangle ABC. The vector a = (radical 2, cos (a + b) / 2, sin (a-b) / 2), and Please, please And the module length of vector a = root 6 / 2. Find Tana times tanb.

1/3
It is known that:
（ √2cos(A+B)/2)^2+ (sin(A-B)/2)^2=(√6/2)^2
cos(A+B)+1+1/2(1-cos(A-B))=3/2
1/2cosAcosB-3/2sinAsinB=0
tanAtanB=sinAsinB/(cosAcosB)=1/3