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In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, if the product of vector AB and vector AC is equal to the product of vector BA and vector BC and is equal to k (k is a real number) (1) Judge the shape of triangle ABC; (2) If c=√2, find the value of k. In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, if the product of vector AB and vector AC is equal to the product of vector BA and vector BC and k (k is real) (1) Judge the shape of triangle ABC; (2) If c=√2, find the value of k.
Since AB.AC=BA.BC, i.e.|AB AC|**cosA=|BA BC|*cosB|AC|* cosA=|BC|*cosB, i.e. AC, BC is projective equal on AB,|AC||=|BC|.
In △ABC, known AB • AC =3 BA• BC. (1) Verification: tanB=3tanA; (2) If cosC = 5 5. Find the value of A.
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In triangle ABC, the opposite sides of angles A, B, C are a, b, c respectively, if vector AB* vector AC = vector BA* vector BC =1 1. Verify A=B 2. Find the value of side length c 3. If |vector AB+vector AC|=root 6, find the area of ABC. In triangle ABC, the opposite sides of angles A, B, C are a, b, c respectively, if vector AB * vector AC = vector BA * vector BC =1 1. Verify A=B 2. Find the value of side length c 3. If |vector AB+vector AC|=root 6, find the area of ABC.
1 Proof: Vector AB * Vector AC = Vector BA * Vector BC =1 Vector AB * Vector AC =- Vector AB * Vector BC Vector AB ×(Vector AC + Vector BC)=0(Vector AC + Vector CB)(Vector AC - Vector CB)=0AC = CBA = B2 Vector AB * Vector AC =1c*b*cosA =1 cosA =(b^2+ c^2-a^2)/(2bc) and a = b...
In triangle ABC, the opposite sides of angles A, B and C are a, b and c respectively, if vector AB* vector AC=vector BA* vector BC=k 1. Judge the shape of triangle ABC 2, C = value of k at root 2
1) Bc cosA=accosB, so cosA/cosB=a/b=sinA/sinB
So sinAcosB-cosAsinB=sin (A-B)=0,
A = B, triangle ABC is isosceles triangle
2) Defined by inner product k=c*(c/2)=c^2/2=1
It is known that the area S of triangle ABC satisfies that the root number 3 is greater than or equal to S is less than or equal to 3, and the vector AB* vector BC=6, with the included angle a (1) Find the value range of a (2) Find the minimum value of f (a)=(sina)^2+2sina*cosa+3(cosa)^2
3 AB||BC|sina/2≤3===>2√3 AB||BC|sina≤6...... (1)
|AB||BC|cosa=6... ...(2)(1)/(2):√3/3≤Tana≤1
30O≤a≤45o
F (a)= cos2a+sin2a=√2[ cos45o cos2a+sin45o ]=√2 cos (2a-45o)
15O≤2a-45o≤45o, where the cosine is a minus function
F (a)=(sina)^2+2sina*cosa+3(cosa)^2 min =√2 cos45o=1
Given that the area of △ABC satisfies: the root number 3/2≤S is less than or equal to 3/2 and the vector AB*BC=3, the included angle between AB and BC is θ,(1) find the value range of the angle of θ,(2) find the maximum value of the function f (θ)=3sin^2 2 times the root number 3sinθcos cos The root 3 is two-thirds higher than the root and not two-thirds lower Given that the area of △ABC satisfies: the root number 3/2≤S is less than or equal to 3/2 and the vector AB*BC=3, the included angle between AB and BC is θ,(1) find the value range of θ angle,(2) find the maximum value of the function f (θ)=3sin^2 2 times the root number 3sinθcos cos Root 3 is two-thirds higher than root 2 Given that the area of △ABC satisfies: the root number 3/2≤S is less than or equal to 3/2 and the vector AB*BC=3, the included angle between AB and BC is θ,(1) find the value range of θ angle,(2) find the maximum value of the function f (θ)=3sin^2θ+2 times the root number 3sinθcos cos Root 3 is two-thirds higher than root 2
Let the opposite sides of angles A, B and C be a, b and c respectively.
(1) If the angle between vector AB and BC is θ, then B=π-θ.
The vector AB*BC=|AB BC|cosB=accos (π-θ)=-accosθ=3, ac=-3/cosθ.
△ABC area =(1/2) acsinθ=(1/2)*(-3/cosθ) sinθ=-(3/2) tanθ.
By title √3/2
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