In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and the area of △ ABC is obtained by satisfying cos (A / 2) = (2 √ 5 / 5) and vector AB * AC = 3 (1) (2) If C = 1, find the value of A

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and the area of △ ABC is obtained by satisfying cos (A / 2) = (2 √ 5 / 5) and vector AB * AC = 3 (1) (2) If C = 1, find the value of A

Cosa = 2 (2 root signs 5 / 5) ^ 2-1 = 2 * 4 / 5-1 = 3 / 5
sinA=4/5
AB*AC=|AB||AC|cosA=3
|AB||AC|=3/(3/5)=5
S(ABC)=1/2|AB||AC|sinA=1/2*5*4/5=2
(2)S=1/2bcsinA=2
1/2b*1*4/5=2
b=5
a^2=b^2+c^2-2bccosA=25+1-2*5*1*3/5=20
A = 2 root 5

In triangle ABC, the angles a, B, C, the opposite sides are a, B, C respectively, and satisfy cos (a \ 2) = ((2 and root sign 5) \ 5)), and the point of vector AB is multiplied by AC = 3 Find the area of triangular row ABC (2 if B + C = 6) and find the value of A If B + C = 6, find the value of A

Cosa = 2cos (A / 2) square - 1 = 0.6, then multiply vector AC = 3bC by cosa = 3bC = 5 according to vector AB (in Sina, you can use sin square + cos square = 1, or you can build a triangle, which will not be discussed in detail here) and then use s = bcsina / 2 (get the first question, area) to ask the second question. Then, it's very stupid. You jointly set the known B + C = 6, and get

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and COS (A / 2) = (2 √ 5) / 5 is satisfied, and vector AB is multiplied by vector AC = 3 (1) Find the area of △ ABC (2) If B + C = 6, find the value of A

cosA=2(cosA/2)^2-1=2*4/5-1=3/5
Then AB * ac * cosa = 3 can be obtained according to the vector
So AB * AC = 5
From cosa > 0, we can know that ∠ A is an acute angle in the triangle, so Sina = 4 / 5
So area = 1 / 2Ab * ac * Sina = 2
AB * AC = 5, i.e. BC = 5
a^2 = b^2 + c^2 - 2bc cosA =(b+c)^2 - 2bc -2bc cosA
=36 - 10 -10x3/5 =20
A = 2x root 5

If the square of vector BC + vector AB multiplied by the point of vector AB in triangular ABC = 0, what is the shape of triangular ABC?

ab*bc+ab*ab=0
=>ab*(ab+bc)=0
=>ab*ac=0 (ab+bc=ac)
=>A is the right angle of the triangle ABC
The above AB represents the vector

In triangle ABC, the square of AB vector is equal to ab vector multiplied by AC vector plus Ba vector multiplied by BC vector plus CA vector multiplied by CB vector, then what triangle is triangle ABC? Please explain the reason,

△ ABC is a right triangle with ab as the hypotenuse. (vector AB) ²＝ Vector ab · vector AC + vector BA · vector BC + vector Ca · vector CB, ∅ vector ab · (vector ab - vector AC) = vector BC · (vector ba - vector CA), ∅ vector ab · (vector ab - vector AC) = vector BC · (vector ac

In triangle ABC, the opposite side of ABC is ABC, which satisfies the square of (AB is the vector) 2Ab * AC = a - the square of (B + C), 1, find the size of angle A 2. Find the maximum value of twice the root sign 3 * cos ^ 2 (C / 2) - sin [(4 Pie / 3) - b], and find the size of angles B and C when obtaining the maximum value

1. According to the meaning of the topic: 2Ab * AC = 2bccosa = a ^ 2 - (B + C) ^ 2
According to the cosine theorem: 2bccosa = B ^ 2 + C ^ 2-A ^ 2, so BC = a ^ 2-B ^ 2-C ^ 2
So: cosa = (b ^ 2 + C ^ 2-A ^ 2) / (2BC) = (b ^ 2 + C ^ 2-A ^ 2) / (2 (a ^ 2-B ^ 2-C ^ 2)) = - 1 / 2
So: a = 2pai / 3 (120 degrees)
2. Original formula = 2gen3 (COSC / 2) ^ 2-sin (4pai / 3-B)
=gen3(1+cosC)-sin(pai+pai/3-B)
Because: B + C = Pai / 3, so:
Original formula = gen3 + gen3cos (PAI / 3-B) + sin (PAI / 3-B)
=gen3+2((gen3/2)cos(pai/3-B)+(1/2)sin(pai/3-B))
=gen3+2sin(2pai/3-B)
=gen3+2sin(pai/3+B)
Therefore, when Pai / 3 + B = Pai / 2, that is, B = Pai / 6, the original formula takes the maximum value of 2 + gen3, and C = b = Pai / 6