In △ ABC, given BC = 2, vector AB times vector AC = 1, then the maximum value of △ ABC area is ■

In △ ABC, given BC = 2, vector AB times vector AC = 1, then the maximum value of △ ABC area is ■

From the question: vector AB * vector AC = [absolute value AB] * [[absolute value AC] cosa = 1, therefore, 1 = AB? AC? Cos? A -------- - (1) because, s = (1 / 2) * [absolute value AB] * [[absolute value AC] Sina, therefore, 4S? = AB? AC? Sin? A --------

Given that the triangle ABC satisfies (all the following are vectors) AB * AB = AB * AC + Ba * BC + ca * CB, then triangle ABC must be A equilateral B oblique C isosceles right angle D right triangle? By the way, what is an hypotenuse triangle?

If AB * AB = AB * AC + Ba * BC + ca * CB, then AB * AB = AB * AC + AB * CB + ca * CB, then AB2 = AB * (AC + CB) + ca * CB, then AB2 = AB * AB + ca * CB, then ca * CB = 0, then the triangle is a right triangle

In the triangle ABC, the opposite side of the angle a B C is a B C, and the vector AB * vector AC = vector ca * vector CB, 1. Judge the shape of ABC 2. Vector ca * vector CB = 8 find B

1. ∵ vector AB * vector AC = vector ca * vector CB ᙽ bccosa = abcosc, that is, ccosa = acosc. From the sine theorem, we can know that: A / Sina = C / sinc, i.e., a = csina / sinc ᙽ ccosa = csina / sinc * COSC, i.e., sinccosa = sinacoscsin (A-C) = 0  a-c = 0, that is, a = C ﹤ ABC is an isosceles triangle

The triangular a, B, C and area s of △ ABC satisfy the relationship: S = C2 - (a-b) 2 and a + B = 2. Find the maximum value of area s

By substituting the cosine theorem C2 = A2 + b2-2abcosc and the area formula s = 12absinc, it is obtained that s = C2 - (a-b) 2 = A2 + b2-2abcosc - (a-b) 2, i.e., 12absinc = 2Ab (1-cosc), ν1 − coscsinc = 14, let 1-cosc = k, sinc = 4K (k > 0), from (1-k) 2 + (4K) 2 = cos2c + sin2c = 1

In the triangle ABC, if the vector AB ^ 2 = AB * AC + Ba * BC + ca * CB, what triangle is this? 1 equilateral triangle 2 acute triangle 3 right triangle 4 obtuse triangle

Option 1
Steps:
AB^2=AB(AC+BC)+AC×BC
AB^2-AB(AC+BC)-AC×BC=0
(AB-AC) (AB + BC) = 0 or (AB + AC) (ab-bc) = 0
So choose 1

In triangle ABC, given vector AB times vector AC = 9, area of triangle = 6, BC = 4, then the circumference of triangle is?

Answer: 12
bcCosA=9.1
S=SinAbc/2=6.2
CosA=b²+c²-a²/2bc.3
Replace 3 with 1
The result shows that B 2 + C 2 - a 2 = 18 A = 4
So B 2 + C 2 = 34
(B + C) 2 - 2BC = 34.4
Add up 1 and 2 squared (this problem is difficult here, the purpose is to use sin? X + cos? X = 1)
BC = 15.5
Substituting 5 into 4 gives B + C = 8
Finally, a + B + C = 12

In the triangle ABC, ab = AC = 5, BC = 6. With a straight line, the circumference and area of the triangle ABC are divided into two equal parts? After dividing, the distance between the intersection point of each line and the ABC edge of the triangle and the vertex of the triangle is also pointed out

This is an isosceles triangle, ab = AC = 5; BC = 6;
The circumference and area of the isosceles triangle ABC can be divided into two equal parts: ad = 4; BD = CD = 3; ad = 4; BD = CD = 3;

Given the circumference of the triangle ABC a + B + C = 6, B ^ 2 = AC, find the maximum area of the triangle ABC, find the Ba vector and click the range of BC vector

CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac ≥ 1 / 2, so B ≤ 60
The equation a + C + √ AC = 6, AC ≤ 4 both hold the equal sign when a = C, so the maximum area of the triangle ABC = acsinb / 2 = √ 3
Vector Ba * BC = accosb = (a ^ 2 + C ^ 2-B ^ 2) / 2 = [(6-b) ^ 2-3b ^ 2] / 2 = - B ^ 2-6b + 18 (0

It is known that the circumference of the triangle ABC is 6, a, B, C. in equal proportion sequence, (1) find the maximum area of triangle ABC; (2) the range of vector Ba * vector BC

B ^ 2 = AC according to the cosine theorem B ^ 2 = a ^ 2 + C ^ 2-2accosb and S = acsinb / 2 plus a + B + C = 6, we can get CoSb = (a ^ 2 + C ^ 2-ac) / 2Ac = A / 2C + C / 2a-1 / 2 according to the basic inequality, the minimum value of CoSb is 1 / 2, so the maximum value of SINB is the root sign 3 / 2, so the maximum area is a = C = 2

It is known that the area s of the triangle ABC satisfies that the root sign 3 is greater than or equal to s less than or equal to 3, and the vector AB * vector BC = 6, and its angle is a (1). Find the value range of a (2) find the minimum value and maximum value of F (a) = (Sina) ^ 2 + 2sina * cosa + 3 (COSA) ^ 2

(1) S / T = s / 6 = 1 / 2 * Sina / cosa = 1 / 2 * Sina / cosa = 6S / T = s / 6 = s / 6 = 1 / 2 * Sina / cosa = 1 / 2tana, s = s = 3tana  s = 3 Tana  s = 3  3 ≤ s ≤ 3, ∵ 3 / 3 ≤ s / 3 = Tana ≤ 1, and \\\\124a) ^ 2 + 2sina * cosa + 3 (COSA) ^ 2 =