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As shown in the figure, the vertex a (- 2 √ 3,0) of the equilateral triangle ABC, B, C are on the Y axis.)
Let the equilateral △ ABC, B be above the x-axis and C below the x-axis,
OA = 2 √ 3 is high, ab = 2 √ 3 × 2 / √ 3 = 4, OB = OC = 2
（1）∴B（0,2）,C（0,-2）.
（2）S△ABC=2√2×4/2=4√2.
Perimeter L = 4 × 3 = 12
As shown in the figure, the vertex a (- 2 √ 3,0) of the equilateral triangle ABC, B, C are on the Y axis.)
B(0,2)
C(0,-2)
C=12
S=4√3
In the triangle ABC, a, B, C are the opposite sides of three inner angles a, B and C. If a = 2, C = π / 4, CoSb / 2 = 2, root sign 5 / 5, the area s of triangle ABC is obtained I don't know why CoSb = 2cos 2 (B / 2) - 1 = 3 / 5, and how this formula comes from
Because the angle doubling formula of trigonometric function:
coa2α=2coa²α-1
In triangle ABC, the opposite sides of a, B and C are a, B, C respectively. We know that C = 2, C = 3. If the area of triangle ABC is equal to the root sign 3, find the value of a and B. (2) if (SINB) square = 2 (Sina) square, find the area of triangle ABC
1) According to the cosine formula: C ^ 2 = a ^ 2 + B ^ 2-2abcosa
Therefore, a ^ 2 + B ^ 2-2abcos (π / 3) = 2 ^ 2 = 4
a^2+b^2-ab=4…… （1）
And the area s = absinc / 2 = √ 3
absin(π/3)=2√3
ab=4…… （2）
From (1) and (2), a = 2, B = 2
2) Because SINB = 2sina, so B = 2A, and a ^ 2 + B ^ 2-2abcosc = C ^ 2, the simultaneous solution gives a = 2 times and 3 / 3, B = 4 times and 3 / 3,
So the area s = 1 / 2absin C = 2 times √ 3 / 3
In the triangle ABC, the opposite sides of a, B and C are a, B, C respectively. We know that C = 2, C = 3. If the area of triangle ABC is equal to the root sign 3, find the values of a and B. (2) if SINB = 2sina, find the area of triangle ABC
Answer: 1) according to the cosine formula: C ^ 2 = a ^ 2 + B ^ 2-2abcosc, therefore: A ^ 2 + B ^ 2-2abcos (π / 3) = 2 ^ 2 = 4A ^ 2 + B ^ 2-AB = 4 (1) And the area s = absinc / 2 = √ 3absin (π / 3) = 2 √ 3AB = 4 (2) From the solutions of (1) and (2), a = 2, B = 22) sinc + sin (B-A) = 2sin2asin (π
In the triangle ABC, the sides of angles a, B and C are a, B and C respectively. We know that a = 2, B = radical 7, and B is equal to the third derivation. Find the value of the edge and the face of the triangle ABC
According to the sine theorem
a/sinA=b/sinB
sinA=asinB/b=√（3/7）
So cosa = √ (4 / 7)
sinC=sin（180-A-B）
=sin（A+B）
=sinAcosB+sinBcosA
=sinA/2+√3/2cosA
=√（3/28）+√（3/7）
=3/2√（3/7）
According to the sine theorem
a/sinA=c/sinC
c=asinC/sinA=3
S triangle ABC = 1 / 2Ac * SINB = 3 / 2 √ 3
In the triangle ABC, sinc COSC + SINB = radical 3, and angle a ≥ quartile, then SINB =? As the title
sinC-cosC=√2sin(C-π/4)=√2cos(3π/4-C)=-√2cos(π/4+C)
Obviously, C > = π / 4, otherwise COSC > = √ 2 / 2, sinc + SINB = √ 3 + COSC > = 2, can not be true
So π / 4 + C > π / 2, - 2cos (π / 4 + C)
In △ ABC, if 3B = 2 If 3asinb and CoSb = COSC, then the triangle must be () A. Isosceles triangle B. Equilateral triangle or isosceles triangle C. Equilateral triangle D. Right isosceles triangle
In △ ABC, if 3B = 2
If 3 asinb and CoSb = COSC, then 3sinb = 2
3 sinasinb, and B = C,
Sina is obtained=
Three
2，∴A=π
3 or 2 π
3．
When a = π
When 3, then from b = C, we can get that △ ABC is an equilateral triangle,
When a = 2 π
When 3, then from b = C, △ ABC is an isosceles triangle,
Therefore, B
In the triangle ABC, a, B, C are the lengths of the opposite sides of the angles a, B and C respectively, and satisfy the requirements of CoSb / COSC = - B / (2a + C), so as to find the value of angle B. If b = radical 13, a + C = 4 Find the area of triangle ABC If B = radical 13, a + C = 4, find the area of triangle ABC
CoSb / COSC = - B / (2a + C) 2acosb + ccosb + bcosc = 0 by using a as the height ah of BC side, we can get csonb + bcosc = a, so 2acosb + a = 0cosb = - 1 / 2B = 120 ° by cosine theorem B ^ 2 = a ^ 2 + C ^ 2-2accosb, i.e. 13 = a ^ 2 + C ^ 2 + AC = (a + C) ^ 2-ac = 4 ^ 2-acac = 3, so the area of triangle ABC = (1 / 2) acsin
In triangle ABC, angles a, B and C satisfy sinc CoSb = (2sina SINB) COSC (1) to find the size of angle C (2) Find the value range of the function y = 2sinb ^ 2-cos2a
sinC cosB=2sinAcosC-sinB cosC
sinC cosB+sinB cosC=2sinAcosC
sin(B+C)=2sinAcosC
Because a = 180 - (B + C) in the triangle
sin(180-A)=2sinAcosC
sinA(1-2cosC)=0
2cosC=1
cosC=1/2
Zero
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