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In the triangle ABC, BC = 2, Sina = 2 / 3, root 2, then what is the maximum product of AB vector and AC vector
Product of AB vector and AC vector = AB length * AC length * cosa = CB * cosa
(1) A is an acute angle, cosa = 1 / 3, the product of AB vector and AC vector = CB * (1 / 3) ---- 1
According to the cosine theorem, a ^ 2 = 4 = B ^ 2 + C ^ 2-2bc * cosa, 2BC - (2 / 3) BC = (4 / 3) BC
Take the equal sign condition, B = C = radical 3
The product of AB vector and AC vector is CB * (1 / 3) 1;
(2) If a is an obtuse angle, cosa = - 1 / 3, the product of AB vector and AC vector = CB * cosa = - (1 / 3) BC
Because a ^ 2 = 4 = B ^ 2 + C ^ 2-2bc * cosa "2BC + (2 / 3) BC = (8 / 3) BC, that is BC" 3 / 2, take the equal sign condition: B = C = (root sign 6) / 2; the product of AB vector and AC vector = CB * (- 1 / 3) (- 1 / 3) BC "- 1 / 2, the maximum value of BC tends to 0, so there is no maximum value at this time
To sum up, the obtuse angle is the maximum;
When a is an acute angle, the maximum value is 1
When △ ABC, B = 60, C = 7, s △ ABC = 21, root 3, then a= As the title
S=acsinB/2
21√3=a×7×√3/2 ÷2
a=12
In the triangle ABC, the angle a is 120 ° C > b, and the angle ABC = root 3. Find B a = root 21 s In the triangle ABC, the angle a is 120 ° C > b, the angle ABC = root 3, find B a = root 21, s ABC = root 3, find B C
According to the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2 * b * c * cosa, we get 21 = B ^ 2 + C ^ 2 + BC (1) according to the triangle area formula: S = 1 / 2bcsina, we can get the following formula: √ 3 = 1 / 2BC * √ 3 / 2, then we can get BC = 4 (2) and (1) (2) two formulas
In the triangle ABC, the angle a is 120 °, CC C is greater than B, a = root 21, the area of triangle ABC is root 3, B C
Area of triangle = radical 3 = bcsina / 2 = bcsin120 / 2
bc=4
b^2+c^2-a^2=2bccosA
b^2+c^2-21=2bccosA=8*(-1/2)=-4
b^2+c^2=17
(b+c)^2=25
b+c=5
c>b,
c=4,b=1
A is a real number, vector AB = (a, 1), vector AC = (3,4). If the length of AB is less than or equal to root 10, what is the probability that the area of triangle ABC is greater than 5?
AB^2=a^2+1
In △ ABC, given vector | ab | = 4, vector | AC | = 1, s △ ABC = radical 3, then vector ab × vector AC is equal to In △ ABC, given vector | ab | = 4, vector | AC | = 1, s △ ABC = radical 3, then vector ab × vector AC is equal to A.-2 B.2 C.± 2 D.± 4
Sine theorem:
S△ABC
=Radical 3
=1/2AB*AC*sinA
=1/2*4*1*sinA
=2sinA
Sina = (radical 3) / 2
Cosa = plus or minus 1 / 2
That is, vector ab × vector AC = AB * AC8 plus or minus 1 / 2
So choose C
In ABC, vector AB = vector a, vector AC = vector B, vector a * vector B is less than 0, triangle area is 15 times, root sign 3 / 4, a module is 3, B module is 5, then BC side length
S=1/2|AB|*|AC|*sinA =1/2*3*5sinA
=15√3/4
The solution is: Sina = √ 3 / 2
Therefore: a = 120 ° or a = 60 °
Because: ab vector dot product AC vector is less than 0
Therefore: A>90 °
Therefore, a = 120 degrees
|Vector BC ^ 2 = vector BC ^ 2 = (vector AC vector AB) ^ 2
= AC^2+AB^2-2AC•AB
=25+9-2*5*3*cosA
=25+9-2*5*3*cos120°
=49,
∴|BC|=7.
Given that the triangle ABC is an acute angle triangle, the corresponding sides of angles a, B and C are a, B, C. the angle B is 45 degrees, B is the root 2, and C is the root 3. Find the area of the triangle ABC
According to the sine theorem: B / SINB = C / sinc
So root 2 / sin45 ° = root 3 / sinc
So sinc = the root of two is 3
Given that the triangle ABC is an acute angle triangle, then C is 60 degrees
So a is 75 ° and Sina = (root 6 + root 2) / 4
Therefore, s = 0.5 * Radix 2 * Radix 3 * (Radix 6 + Radix 2) / 4 = (3 + Radix 3) / 4
It is known that a, B and C are the opposite sides of angle a, angle B and angle C in the acute triangle ABC. If a = 3, B = 4, and the area of the triangle is 3, then C=
Using the formula: S = (1 / 2) * AB * sinc
Sinc = ～
COSC = 1 / 2
Cosine theorem: square of C = a + B - 2abcosc
So C = heel 13
In △ ABC, a = π / 6, (1 + radical 3) C = 2B, find the angle c, if the vector CB multiplies the vector CA = 1 + the root 3, find the edge a.b.c
(1) By sine theorem
(1+√3)/2=b/c=sinB/sinC=sin(120°-C)/sinC=(√3/2*cosC+1/2*sinC)/sinC
∴tanC=1
∴C=45°
(2)
∵CB/CA=sinA/sinB=sin60°/sin75°=2√3/(√6+√2)
∵ vector CB * vector CA = CB * ca * COSC = 1 + √ 3
∴CB*CA=√6+√2
∴CB^2=2√3
That is, CB = 4 times √ 12 = a
In addition, B = 4 times √ 3 + 1 / (4 times √ 3) and C = 2 / (4 times √ 3) are obtained correspondingly
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