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Given vector m=(sinA, cosA), n=(root number 3,-1), m*n=1, and A is an acute angle, find the size of angle A
M*n=√3 sinA-cosA=1.
2(√3SinA-1/2cosA)=1.
2(SinAcos30°-cosAsin30°)=1.
Sin (A-30°)=1/2
Sin (A-30°)= sin 30°.
A-30°=30°.
A =60°
M*n=√3 sinA-cosA=1.
2(√3SinA-1/2cosA)=1.
2(SinAcos30°- cosAsin30°)=1.
Sin (A-30°)=1/2
Sin (A-30°)= sin 30°.
A-30°=30°.
A =60°
Given vector OA=(3.-4), vector OB=(6.-3), vector OC=(5-m,-3-m). If ∠ABC is an acute angle, is the value range of the number m? Given vector OA=(3.-4), vector OB=(6.-3), vector OC=(5-m,-3-m).
Vector BA=vector OA-OB=(-3,-1)
Vector BC=vector OC-OB=(-1-m,-m)
Because acute angle 1> Cos >0
That is, BA point multiplication BC is greater than 0
1>3+M+m >0
-3<2M <-2
-1.5
Given the vector OA vector =(3,-4), OB vector =(6,-3), OC vector =(5-m,-3-m),(1) if points A, B and C can form a triangle, the condition that the number m should be satisfied;(2) if triangle ABC is a right triangle, the value of the number m should be obtained
Suppose that the points A, B, C can not form a triangle, i.e., A, B, C are collinear, then let AC=xAB,(x is a real number), i.e.,(-M-22,-M-33)=x (-21,-39), then -M-22=-21x-M-33=-39x, then the solution x=11/18, M=-55/6, then the three points A, B, C are not collinear if x=11/18, M=-55/6, i.e., points A, B, C can form a triangle, real...
Given vector OA=(3,-4) vector OB=(6,-3) vector OC=(5-M, M-3) 1) If A, B, and C can form a triangle, the condition that should be satisfied is that the number m should be obtained. 2) If △ABC is a right triangle, obtain the value of m. Given vector OA=(3,-4) vector OB=(6,-3) vector OC=(5-M, M-3) 1) If A, B, C can form a triangle, the condition that should be satisfied by the fact that the number m. 2) If △ABC is a right triangle, obtain the value of m.
0
As shown in the figure, in the parallelogram ABCD, E and F are the midpoint of BC and DC respectively, and G is the intersection point. AB = A, AD = B. Try A, B is the base representation CG ______div class=" sanwser ">−1 3( A+ As shown in the figure, in the parallelogram ABCD, E and F are the midpoint of BC and DC respectively, and G is the intersection point, if AB = A, AD = B. Try A, B is the base representation CG ______div class=" sanwser ">−1 3( A+ As shown in the figure, in the parallelogram ABCD, E and F are the midpoint of BC and DC, and G is the intersection point, if AB = A, AD = B. Try A, B is the base representation CG ______div class=" sanwser ">−1 3( A+
G is the center of gravity of △BCD,
Gu
CG =1
3
CA=-1
3
AC=-1
3(
AD+
AB)=-1
3(
A+
B)
Therefore, the answer is -1
3(
A+
B)
G is the center of gravity of △BCD,
Gu
CG=1
3
CA=-1
3
AC=-1
3(
AD+
AB)=-1
3(
A+
B)
Therefore, the answer is -1
3(
A+
B)
Given Point A (-1,5) and Vector A =(2,3) if AB =3 A, the coordinates of point B are () A.(7,4) B.(7,14) C.(5,4) D.(5,14)
0
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