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In triangle ABC, abc is the opposite edge of angle ABC, vector m=(2b-c, cosC), vector n=(a, cosA) m parallel n, find the magnitude of angle A.
From vector M parallel vector N:
(2B-c)/a=cosC/cosA
Substitute sine theorem: sinA=a/2r, sinB=b/2r, sinC=c/2r
(4RsinB-2rsinC)/2rsinA = cosC / cosA 2r can be eliminated at this time
(2SinB-sinC)/sinA = cosC/cosA
2SinBcosA=sinAcosC+sinCcosA=sin (A+C)=sinB2cosA=1, i.e. A=60°, is obtained by cross multiplication of 2sinBcosA=sinsinAcosC+sinCcosA=sin (A+C)=sinB with both sides removed
The opposite sides of the three angles of a triangle are abc.m vector=(2b-c, a) n vector=(cosA,-cosC) and m vector is perpendicular to n vector. When Y =2 times of sin square B + sin (2B +6 division) takes the maximum value, calculate the angle B
If the vector is vertical, then there is cosA (2b-c)-acosC=0, i.e. sinAcosC=cosA (2sinB-sinC), and the reduction is sin (A+C)=2cosAsinB, so cosA=1/2, A=60°.
If the vector is vertical, then there is cosA (2b-c)-acosC=0, i.e. sinAcosC=cosA (2sinB-sinC), and reduction is sin (A+C)=2cosAsinB, so cosA=1/2, A=60°.
In triangle ABC, angles A, B, C are opposite sides of a, b, c. The known vectors m =(2b-c, a), n =(cosA,-cosC), and m is perpendicular to n, (1) Find the size of angle A,(2) If a = root number 3, the area of triangle ABC is triple root number 3/4, try to determine the shape of triangle ABC and explain the reason In triangle ABC, angles A, B, C are opposite sides of a, b, c, respectively. (1) Find the size of angle A,(2) If a = root number 3, the area of triangle ABC is triple root number 3/4, try to determine the shape of triangle ABC and explain the reason
(2B-c) cosA=acosC
A=2RsinA b=2RsinB c=2RsinC
(2SinB-sinC) cosA = sinAcosC
2SinBcosA-sinCcosA=sinAcosC
2SinBcosA = sin (A+C)= sinB
SinB=0 2 cosA=1 A=60°
In triangle abc, vector m=(c-2b, a) vector n=(cosa, cosc) and vector m vertical vector n is known Find the size of angle a if vector ab is multiplied by vector ac=4 to find the minimum value of b c side
Vector m vertical vector n, then: m*n=0, get:
(C-2b) cosA+acosC=0
(SinC-2sinB) cosA+sinAcosC=0
CosAsinC+sinAcosC=2sinBcosA
Sin (A+C)=2sinBcosA
SinB=2sinBcosA
CosA=1/2
Get: A=60°
AB*AC=bccosA=4, then: bc=8
Again: a2= b2+ c2-2 bc cosA = b2+ c2- bc
Because b2+c2≥2bc, then: b2+c2≥16, so:
A2+bc=b2+c2≥2bc
A2≥ bc =8
Then: a = the minimum value of BC is 2√2
In triangle ABC, the opposite sides of A, B, C are a, b, c. Let vector m=(a, c), vector n=(cosC, cosA) (1) If m//n, c=root number 3a, find the angle A. (2) If m*n=3bsinB,cosA =4/5, find the value of cosC
(1) From m//n, acosA=ccosC and then sinAcosA=sinCcosC, so sin2A=sin2C and then 2A=2C (closed) or A+C=90 degrees, from c=root3a, so sinC=root3sinA and then sin (90 degree-A)=cosA=root3sinA and then A=30 degrees (2) m*n=3bsinB, so acosC+ccosA=b=3bsinB, so sin...
In triangle ABC, given BC=2, vector AB * vector AC=1, then the maximum value of triangle ABC area is? In triangle ABC, if BC =2 and vector AB * vector AC =1, then the maximum value of triangle ABC area is?
AB*AC=/AB/*/AC/cosα=(AB^2+AC^2-BC^2)/2=1
So AB^2+AC^2=6≥2/AB/*/AC/
/AB/*/AC/≤3, so cos 1/3
So sin 2√2/3 S=0.5*/AB/*/AC/sin 2
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