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

In the triangle ABC, the edges of angles a, B and C are a, B, C respectively, and satisfy cosa / 2 = 2, root sign 5 / 5, vector AB * 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)
Cosine theorem a ^ 2 = B ^ 2 + C ^ 2-2cosa * BC obtains a = 2 * radical 5

Let a, B, C be the inner angles of △ ABC, vector a = (COS (a-b) / 2, Radix 3sin (a + b) / 2), and | a | = radical 2 if C When the maximum is, there is a moving point m such that | Ma |, | ab |, and | MB | are isometric series, then the maximum value of | MC | / | ab | is

a^2=[cos(A-B)/2]^2+[√3sin(A+B)/2]^2
=(1/2)[1+cos(A-B)+3-3cos(A+B)]=2,
∴0=cos(A-B)-3cos(A+B)=cos(A-B)+3cosC,
When C is maximum, a = B, COSC = - 1 / 3,
|Ma |, | ab | and | MB | are isometric series,
<==>|MA|+|MB|=2|AB|,
The trajectory of M is an ellipse with a, B as the focus and 2 | ab | as the major axis
The ratio has nothing to do with the choice of units, so let | ab | = 2, the midpoint of AB is O, and a = b point | AC | = | BC | = P,
According to the cosine theorem, 2p ^ 2 (1 + 1 / 3) = 4, P ^ 2 = 3 / 2,
∴|OC|=√(p^2-1)=1/√2,
If M is the endpoint of the minor axis of the above ellipse (on both sides of AB with point C),
Then | om | = √ 3 (if you want to demonstrate, you need to establish a coordinate system),
|The maximum value of MC | / | ab | is (1 / √ 2 + √ 3) / 2 = (√ 2 + 2 √ 3) / 4

In the triangle ABC, the angles a, B, C satisfy that CSA / 2 is two fifths times the root sign five, the vector AB times the vector AC equals 3, find the area of the triangle

CSA / 2 should be wrong, is it cos A / 2?
From the vector AB multiplied by the vector AC equals 3,
AB*AC*cosA=3
But cos A / 2 = 2 √ 5 / 5
Therefore, cosa = 2 (COS A / 2) ^ 2 - 1 = 0.6 can be obtained
So AB * AC = 5,
From cosa = 0.6, sina is equal to 0.8
So the area of the triangle is 0.5 × ab × AC × Sina
=2

In the triangle ABC, the sides of the angle A.B.C are a.b.c. and cosa / 2 = 2 times root sign 5 / 5, vector AB * vector AC = 3, the area of triangle ABC is calculated If C = 1, find the value of A

Look at the picture

In △ ABC, the edges corresponding to angles a, B and C are a, B and C respectively, and satisfy cosa 2=2 Five 5， AB• AC=3． (I) calculate the area of △ ABC; (II) if B + C = 6, find the value of A

(I) because of cosa
2＝2
Five
5，∴
cosA＝2cos2A
2−1＝3
5，sinA＝4
5，
By
AB•
AC＝3，
Bccosa = 3, ﹥ BC = 5,
∴S△ABC＝1
2bcsinA＝2
(II) for BC = 5 and B + C = 6,
/ / b = 5, C = 1 or B = 1, C = 5,
According to the cosine theorem, A2 = B2 + c2-2bccosa = 20, ν a = 2
Five

In ABC, a = 3, B = 3

Sine theorem:
a/sinA=b/sinB
sinA=asinB/b=√3/2
Zero

In the triangle ABC, if AB = 3, AC = 2, BC = root 10, then AB vector multiplied by AC vector =?

Cosine theorem for angle a
cosA=(9+4-10)/(2*3*2)=1/4
AB vector multiplied by AC vector = = 3 * 2 * cosa = 3 / 2

In the triangle ABC, Sina + cosa = root two, root three * cos = - radical two * cos (π - b), find the inner angle of three

Sina + cosa = root 2, the square of both sides can get sin? A + 2sinacosa + cos? A = 2, so sin2a = 1, so 2A = 90
A=45°
Then, because I don't know what root three * cos = - radical two * cos (π - b), Radix 3 times cos or what, I can't do it

Given the triangle ABC. A = 3 root sign 3 – 4, and cosa = 3 / 5, cos (a + b) = - 1 / 2, find Sina, find the area s in triangle ABC

cos(A+B)=cosC=-1/2 sinC=√3/2
sinA=4/5 a/sinA=c/sinC c=(45-20√3)/8 sinB=sinAcosC+sinCcosA= (3√3-4)/10
S=1/2acsinB=1/2*(3√3-4)*(45-20√3)/8*(3√3-4)/10

In △ ABC, ∠ a, ∠ B are all acute angles, and Sina = 1 2，tanB= 3, ab = 10, find the area of △ ABC

∵ in ᙽ ABC, ᙽ A and ᙽ B are acute angles, Sina = 1
2，tanB=
3，
∴∠A=30°，∠B=60°，∠C=90°，
∵sinA=a
C=1
2tanB=b
A=
3AB=10，
∴a=1
2c=5，b=
3a=5
3，
∴S△ABC=1
2ab=1
2×5×5
3=25
Three
2．