In the acute triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and the root number 3B = 2asinb is known to calculate the size of the angle; If a = 6, find the range of B + C

In the acute triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and the root number 3B = 2asinb is known to calculate the size of the angle; If a = 6, find the range of B + C

From sine: asinb = bsina
∵√3b=2asinB
∴√3b/2=asinB=bsinA
∴sinA=√3/2
∵ a triangle is an acute triangle
∴A=60
∴cosA=1/2
∵a/sinA=b/sinB=c/sinC
∴b=asinB/sinA=4√3sinB
c=asinC/sinA=4√3sinC
∵sinB=sin（A+C)=sin(60+C)
∴sinB=√3/2cosC+1/2sinC
∴b+c=4√3（sinB+sinC)
=4√3(√3/2cosC+1/2sinC+sinC)
=6cosC+6√3sinC
=12sin(C+π/6)
∵A=60
∴C∈(0,2π/3)
∴C+π/6∈（π/6,5π/6）
∴sin(C+π/6)∈(1/2,1]
∴b+c∈（6,12]
If you don't understand anything, you are welcome to ask

△ in ABC, 2asinb = root number 3 * B ∠ a = 60 or 120. If a = 2, the area of triangular ABC is root number 3, find B, C △ in ABC, 2asinb = root number 3 * B, ∠ a = 60 or 120. If a = 2, the ABC area of the triangle is root number 3, find B and C

sinA=√3/2
Then: S = (bcsina) / 2 = √ 3
Available: BC = 4
By cosine theorem:
(1) When a = 60 °,
a ²= b ²+ c ²- 2bccosA
Namely: a ²= (b+c) ²- 2bc-2bccosA
Namely: 4 = (B + C) ²- 8-4
Get: B + C = 4
BC = 4
So, B = C = 2
（2）A=120°
a ²= b ²+ c ²- 2bccosA
Namely: a ²= (b+c) ²- 2bc-2bccosA
Namely: 4 = (B + C) ²- 8+4
Get: B + C = 2 √ 2
BC = 4
Think of B, C as equation X ²- 2 √ 2x + 4 = 0 root
It is easy to get that the equation has no solution
So, B = C = 2

It is known that in the acute triangle ABC, the opposite sides of the three inner angles a, B and C are a, B and C respectively, if the root sign 3B = 2asinb 1) Find the size of angle A 2) If a = root 7 and C = 2, find the length of side B and the area of triangle ABC

Asinb = (root 3) / 2 * b = bsina, so Sina = (root 3) / 2, that is, a = 60 ° and SINB = (root 3) / 2 * B / A, so CoSb = (root (1 - (3b ^ 2) / (4a ^ 2)) / (2a) C = acosb + bcosa = (root (1 - (3b ^ 2) / (4a ^ 2)) / 2 + B / 2 is substituted into the value of a and C, and the solution is b = 1 or B = - 3, and b > 0, so B = 1 triangular ABC

In acute triangle ABC, the side lengths of angles a and B are a and B respectively. If 2asinb = root 3b, angle a is equal to

Because 2asinb = √ 3b, asinb / b = √ 3 / 2. Because a / b = Sina / SINB, asinb / b = sinasinb / SINB = Sina, Sina = √ 3 / 2. Because triangle ABC is an acute triangle, a = 60 degrees

In an acute triangle, 2asinb = root three B. If a = 6. B + C = 8, find the area of triangle ABC In the second question, why can't we do this to find the reason? (the first question a is 60 degrees) a ²= b ²+ c ²- 2bccosA=36 （b+c） ²- 2bc-2bccosA=36 64-bc=36 bc=28 It seems wrong here. BC should be equal to 3 / 28 What's wrong

a ²= b ²+ c ²- 2bccosA=36
（b+c） ²- 2bc-2bccosA=36
(b+c) ²- 2bc-2bc × 1/2=36
64-3bc=36
3bc=28

In △ ABC, a = 1, B = 45 °, s △ ABC = 2, then the diameter of the circumscribed circle of △ ABC is () A. 5 two two B. 5 C. 5 two D. 6 two

∵ in △ ABC, a = 1, B = 45 °, s △ ABC = 2,
∴1
2acsinb = 2, i.e. C = 4
2，
Obtained from the cosine theorem: B2 = A2 + c2-2accosb = 1 + 32-8 = 25, i.e. B = 5,
Then from the sine theorem, d = B
sinB=5
2．
Therefore: C