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If A, B are the two internal angles of the acute triangle ABC... a mathematical solution ~~ 1. If A and B are two internal angles of acute triangle ABC, in which quadrant is the point P (cosB-sinA, sinB-cosA)? Why? 2. Given that the coordinate of the point P on the final edge of the acute angle α is (2sin2,-2cos2), then α is equal to? 3. Given f (sinx)= sin (4n+1) x, find f (cosx)=? Three questions for answers
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The opposite sides of the internal angles A, B, C in the acute triangle ABC are known to be a, b, c, defining the vector m =
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In the acute triangle ABC, the edges of the internal angles A, B, C are known to be a, b, c, respectively, vector m=(2sin (A+C), root number 3), n=(cos2B,2cos square B/2-1) And vector m, n is collinear (1) Find the size of angle B;(2) if b=1, find the maximum area of triangle ABC In an acute triangle ABC, the edges of internal angles A, B, C are known to be a, b, c, respectively, vector m=(2sin (A+C), root number 3), n=(cos2B,2cos square B/2-1) And vector m, n is collinear (1) Find the size of angle B;(2) if b=1, find the maximum area of triangle ABC
(1)
M, n collinear
=>
2Sin (A+C)/√3= cos2B/(2(cos (B/2))^2-1)
2Sin (A+C).cosB=√3 cos2B
2SinBcosB =√3 cos2B
Sin2B =√3 cos2B
Tan2B =√3
B =π/6
(2)
B=1
By line-rule
A/sinA = b/sinB
A =(b/sinB) sinA
=2SinA
And
C=2sinC=2sin (5π/6-A)
A1= area of ABC
=(1/2) AcsinB
=(1/2)(2SinA)(2sin (5π/6-A))(1/2)
= SinAsin (5π/6-A)
=(1/2)(Cos (2A-5π/6)- cos5π/6)
=(1/2)(Cos (2A-5π/6)3/2)
Max A1 at cos (2A-5π/6)=1
Max A1=(1/2)(1 3/2)
The opposite sides of the internal angles A, B and C in the acute angle △ABC are known as a, b and c respectively, the vector m=(2sinB, root number 3), the vector n=(2cos^2B/2-1, cos2B), and m⊥n,(1) find the size of B (2) If b=2, find the maximum area of △ABC
Vector m=(2sinB, root number 3), vector n=(2cos^2B/2-1, cos2B),
And m⊥n
M ●n=0
I.e.2sinB (2cos2B/2-1)3 cos2B=0
2 Cos2B/2-1= cosB
2SinBcosB 3 cos2B=0
Sin2B 3 cos2B=0
Sin2B =-√3 cos2B
Tan2B=-√3
In △ABC
2B =120o, B =60o
(2)
B=60o, b=2
According to cosine theorem
B2= a2+c2-2 accosB
4= A2+ c2- ac
A2+c2≥2ac
4= A2+ c2-ac ≥ ac
Ac≤4
SΔABC=1/2 acsinB=√3/4*ac 3
The maximum area of ABC is √3
Let A, B, C three internal angles A, B and C of an acute triangle ABC be a, b and c respectively, and let the vector m=(1, sinA 3 cosA), n=(sinA,3/2), and m//n, if a=2, c=4√3 sinB, and the area of the triangle ABC is less than √3, determine the value range of the angle B. Let A, B, C three internal angles A, B and C of an acute triangle ABC be a, b and c respectively, the known vector m=(1, sinA 3 cosA), n=(sinA,3/2), and m//n, if a=2, c=4√3 sinB, and the area of the triangle ABC is less than √3, then the value range of the angle B is obtained. Let A, B, C three internal angles A, B, C of the acute triangle ABC be a, b, c, the known vector m=(1, sinA 3 cosA), n=(sinA,3/2), and m//n, if a=2, c=4√3 sinB, and the area of the triangle ABC is less than √3, find the value range of the angle B
SinA /1=(3/2)/(sinA 3 cosA)
2SinA^2+2√3sinAcosA=3
√3Sin 2A =2+ cos2A
3Sin2A^2=4+4cos2A+cos2A^2
Cos2A=-1/2,2A=2/3π, A=π/3
AH=AB*sinB=c*sinB=4√3sinB^2
S=aAH/2=2*4√3sinB^2/2=4√3sinB^2>=√3
π/6
In the known acute angle △ABC, the three internal angles are A, B, C, two vectors P=(2-2sinA,cosA + sinA), Q=(sinA-cosA,1+sinA), if P vs. Q is a collinear vector. (1) Find the size of ∠A; (2) Find function y=2sin2B+cos (C−3B 2) When the maximum value is taken, the size of ∠B.
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