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Any line passing through the center of gravity of △ABC intersects AB, AC at point D, E. If AD=x AB, AE=y AC, xy=0, then 1 X+1 The value of y is ______.
G is the center of gravity of △ABC
Take straight line DE of G parallel BC
∵
AD=x
AB,
AE=y
AC,
X=2
3, Y=2
3
Then 1
X+1
The value of y is
=3
2+3
2=3
Therefore, the answer is:3
G is the center of gravity of △ABC
Straight line DE through G parallel BC
∵
AD=x
AB,
AE=y
AC,
X=2
3, Y=2
3
Then 1
X+1
The value of y is
=3
2+3
2=3
Therefore, the answer is:3
G is the center of gravity of △ABC
Take the straight line DE of G parallel BC
∵
AD=x
AB,
AE=y
AC,
X=2
3, Y=2
3
Then 1
X+1
The value of y is
=3
2+3
2=3
Therefore, the answer is:3
In triangle ABC, AB =4, BC =2 root number 2, and BA vector multiplied by BC vector =-8, then what is AC = In triangle ABC, AB =4, BC =2, and BA vector multiplied by BC vector =-8, then what is AC =
CosB=(vector BA*vector BC)/(|AB BC|)=-8/[4*(2√2)]=-1/√2 By cosine theorem: AC2=AB2+BC2-2AB*BC*cosB=42+(2√2)2-2*4*(2√2)*(-1/√2)=16+8+16=40 AC=√40=2√10
CosB=(vector BA*vector BC)/(|AB||BC|)=-8/[4*(2√2)]=-1/√2 By cosine theorem: AC2=AB2+BC2-2AB*BC*cosB=42+(2√2)2-2*4*(2√2)*(-1/√2)=16+8+16=40 AC=√40=2√10
In a triangle, cosA/2=2 root number 5/5, vector AB*vector AC=3, find the area of ABC; if AC+AB=6, find the value of BC In a triangle, cosA/2=2 root number 5/5, vector AB * vector AC =3, find the area of ABC; if AC + AB =6, find the value of BC
First question:
Cos (A/2)=2√5/5,[ cos (A/2)]^2=4/5, cosA=2[ cos (A/2)]^2-1=3/5>0,
A is an acute angle, sinA=√[1-(cosA)^2]=√(1-9/25)=4/5.
CosA=vector AB·vector AC/(|vector AB||vector AC|), cosA=3/5,
Vector AB·vector AC/(|vector AB||vector AC|)=3/5,3/(AB×AC)=3/5, AB×AC=5.
Area of ABC=(1/2) AB×ACsinA=(1/2)×5×(4/5)=2.
Second question:
By cosine theorem, there are:
BC^2=AB^2+AC^2-2AB×ACcosA=(AB+AC)^2-2AB×AC-2AB×ACcosA
=36-2×5-2×5×(3/5)=26-6=20.
BC=2√5.
Let M be a point in the triangle ABC, and the vector AB*vector AC=(2 roots 3), angle BAC=30 degrees, define f (M)=(m, n, p), where M, n, p are the area of triangle MBC, triangle MCA, triangle MAB respectively. If f (P)=(0.5, x, y), find the minimum value of 1/x+4/y, and send the graph. Let M be a point in the triangle ABC, and the vector AB * vector AC=(2 roots 3), angle BAC=30 degrees, define f (M)=(m, n, p), where M, n, p are the area of triangle MBC, triangle MCA, triangle MAB respectively. If f (P)=(0.5, x, y), find the minimum value of 1/x+4/y, and send the graph.
0
A, B, C are known as triangle ABC triangulation, vector m=(-1, root 3), n=(cosA, sinA), and m*n=1 Find angle A If (1+sin2B)/(cos^2(B)-sin^2(B))=-3, find tanC..
The vector m=(-1, root 3), n=(cosA, sinA), and m*n=1, so (-cosA+root 3sinA)=0 is simplified by the formula sin (A-π/6)=0, and A=π/6 is obtained according to (1+sin2B)/(cos^2(B)-sin^2(B))=-3, and (cosB+sinB)=-3(cosB-sinB) is simplified by obtain tanB=2 and then t...
In triangle ABC, BC=2, AC=2, AB=3+1, let the outer center of triangle ABC be O, if vector AC=m vector AO+n vector AB, find the value of m, n N is a number, AC, AO, AB is a vector, m is multiplied by vector AO, In triangle ABC, BC=2, AC=2, AB=3+1, let the outer center of triangle ABC be O, if vector AC=m vector AO+n vector AB, find the value of m, n N is the number, AC, AO, AB is the vector, m is multiplied by the vector AO,
By AC=mAO+nAB,
AB•AC = mAB•AO + nAB•AB and
AC•AC=mAC•AO+nAC•AB
(Now only AB•AO and AC•AO are required to obtain the univariate quadratic equation of mn)
Because O is the outer center of the triangular ABC,
So from the cosine theorem,
AB•AO=|AB AO|corner BAO
=1/2(Root 3+1)^2
Similarly, AC•AO=1,
Solve m=-root 3-1, n=root 3
(^.^) Lord ah ~ You see I have to call up so hard, how to give a satisfactory answer ~
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