In triangle ABC, a, B and C are the opposite sides of angle a, angle B and angle C respectively. If 2B = a + C, angle B = 30 °, the area of triangle ABC is 1 / 2, then B is?

In triangle ABC, a, B and C are the opposite sides of angle a, angle B and angle C respectively. If 2B = a + C, angle B = 30 °, the area of triangle ABC is 1 / 2, then B is?

S Δ ABC=acsin30º/2=1/2 ∴2ac=4cosB=(a ²+ c ²- b ²)/ 2ac=[(a+c) ²- 2ac-b ²]/ 2ac=(3b ²- 2ac)/2ac∴√3/2=(3b ²- 4)/4∴3b ²- 4=2√3==>b ²= (√3+1) ²/ 3∴b=√3/3+1...

It is known that the perimeter of triangle ABC is root 2 + 1, and Sina + SINB = root 2sinc. If the area of triangle is 1 / 6sinc, the degree of angle c is

sinA+sinB=√2sinC
sinA/sinC+sinB/sinC=√2
a/c+b/c=√2
(a+b)/c=√2
a+b=√2c
Perimeter is root 2 + 1
a+b+c=√2+1
√2c+c=√2+1
C = 1, i.e. AB = 1
a+b=√2+1-c=√2
The area of triangle ABC is 1 / 6sinc
1/2absinC=1/6sinC
ab=1/3
cosC=(a^2+b^2-c^2)/2ab
={(a+b)^2-2ab-c^2}/(2ab)
={(√2)^2-2*1/3-1^2}/(2*1/3)
={2-2/3-1}/(2/3)
=1/2
C=60°

The area of △ ABC is known to be 2 3. BC = 5, a = 60 °, then the perimeter of △ ABC is __

∵ △ the area of ABC is 2
3，A=60°，
∴1
2AC•ABsin60°=2
3. AC • AB = 8
According to the cosine theorem, BC2 = ac2 + ab2-2ac • abcos60 °
Ac2 + ab2-ac • AB = (AC + AB) 2-3ac • AB = BC2 = 25
‡ (AC + AB) 2-24 = 25, get (AC + AB) 2 = 49, get AC + AB = 7
Therefore, the perimeter of △ ABC AB + AC + BC = 7 + 5 = 12
So the answer is: 12

In triangle ABC, a = 120 °, a = root 21, the area of the triangle is root 3, and the perimeter of the triangle is calculated

From the area formula: 1 / 2bcsina = √ 3, substitute the degree of ∠ a, and simplify it to BC = 4 － ①
From the cosine theorem: B ²＋ c ²- 2bccosA=a ², By substituting known numerical values
b ²＋ c ²＋ bc＝21－－－－－②
(B + C) from ② ²－ bc＝21－－－－－③
① Substitute ③ and get (B + C) after deformation ²＝ twenty-five
So B + C = 5, a + B + C = 5 + √ 21

In △ ABC, a, B and C are the opposite sides of angles a, B and C respectively. If 2B = a + C and B = 30 °, the area of △ ABC is 3 2, then   b=（　　） A. 1+ three B. 1+ three two C. 2+ three two D. 2+ three

∵ B = 30 °, and the area of △ ABC is 3
2，
∴S＝1
2acsin30°＝1
two × one
2ac＝3
2，
That is, AC = 6,
∵2b=a+c，
∴4b2=a2+c2+2ac，①
Then B2 = A2 + C2 − 2Ac is obtained from the cosine theorem ×
three
2，②
The two formulas are subtracted to obtain 3B2 = 2Ac + 2Ac ×
three
2＝12+6
3，
That is, B2 = 4 + 2
3，
That is, B = 1+
3，
Therefore: a

In triangle ABC, a = 2 √ 2. B = 45 degrees, and the area of the triangle is equal to 1, then Sina=

You first calculate C according to acsinb whose area is equal to half of that, then calculate B according to the cosine theorem, and then calculate Sina according to the sine theorem. Although it is a little troublesome, as long as you calculate carefully, you will be able to calculate the result. I hope it will be helpful to you