Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained.

Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained.

TanC = sinC / cosC =3√7>0, so C

Solve the triangle and vector combination. Given that a, b, c are opposite sides of three internal angles A, B, C of ΔABC, vector m (root number 3,-1), vector n (CosA, SinA), if vector m, vector n, and aCosB+bCosA=cSinC, then angle B=? ) Solve the combination of triangle and vector. Given that a, b, c are opposite sides of three internal angles A, B, C of ΔABC, vector m (root number 3,-1), vector n (CosA, SinA), if vector m, vector n, and aCosB+bCosA=cSinC, then angle B=? )

In a triangle, all angles belong to (0,π)
AcosB+bcosA=csinC
According to the sine theorem, a=2R·sinA, b=2R·sinB, c=2R·sinC can be substituted
2R·sinA·cosB+2R·sinB·cosA=2R·sinC·sinC
SinA·cosB+sinB·cosA=sin2C
Sin (A+B)= sin2C
Sin (180-C)= sin2C
SinC = sin2C
Solution: sinC=0(rounding) or sinC=1, so C=π/2
M n
M·n=0
I.e.√3 cosA-sinA=0
2Sin [A-(π/3)]=0
A=π/3
To sum up,∠B=π/6

Solving the Problem of Triangular Vector Combination Please be sure to give me a complete process and train of thought ~ because I don't understand ===[ If you give a brief answer === points can not be given.] △ In ABC, angles A, B, and C are opposite sides of a, b, and c respectively, and 2sin^2 A+B/2(this is connected, so long as you write it on the paper, you will understand what I say, I deliberately write the opening point))+cos2C=1[ i.e.2sin^2A+B/2+cos2C=1] 1. Find the size of angle C. 2. If vector m=(3a, b), vector n=(a,-b/3), vector m⊥ vector n,(m+n)·(m-n)=16, find the value of a, b, c. (M, n are all vectors! "·" Multiply points by the product of quantities.) I can't think of a better way to dial. Especially to solve the triangle this part is very unskilled,. )

Pi is the radian angle, equivalent to 180 degrees 2sin^2[(A+B)/2]+cos2C=1 Here, sin [(A+B)/2]=cos (C/2) Because (A+B)/2=pi/2-C/2sin [(A+B)/2]=sin (pi/2-C/2)=cos (C/2)(simplified) The original formula is 2cos^2(C/2)+2cosC^2-1=12cosC^2+cosC -1=0cos...

A Problem of Solving the Combination of Triangle and Vector In triangle ABC, H is the vertical center, the dot product of vector BH and vector BC is 6, sum of squares of sinA and sinC = square of sinB + sinA * sinC Find:(1) Angle B (2) Determine the shape of triangle ABC when the radius R of triangle ABC is the smallest. A Problem of Solving the Combination of Triangle and Vector In triangle ABC, H is the vertical center, the dot product of vector BH and vector BC is 6, sum of squares of sinA and sinC = square of sinB + sinA * sinC Find:(1) Angle B (2) Determine the shape of triangle ABC when the radius R of the triangle ABC is the minimum.

Sum of squares of sinA and sinC = square of sinB + sinA * sinC
A^2+ c^2= b^2+ac
Comparison with Cosine Theorem cosB=1/2, B=60
The angle between vector BH and vector BC is 30 degrees

Sum of squares of sinA and sinC = square of sinB + sinA * sinC
A^2+ c^2= b^2+ac
Comparison with cosine theorem cosB=1/2, B=60
The angle between vector BH and vector BC is 30 degrees

It involves solving triangles and vectors. In △ABC, the opposite sides of ∠A, B, C are a, b, c respectively, vector m=(a^2, b^2), vector n=(tanA, tanB), and vector m// vector n, then why is the shape of △ABC?

Vector m=(a^2, b^2), vector n=(tanA, tanB), vector m//vector n a2tanB=b2tanA according to sine theorem a=2RsinA, b=2RsinB sin2A*sinB/cosB=sin2B*sinA/cosA sinA/cosB=sinB/cosA sinAcosA=sinBcosB sin2A=sin2...

Vectors and Triangles O is a fixed point on the plane, A, B, C are three non-collinear points on the plane, and the moving point P satisfies OP=OA (AB/|AB AC/|AC|).[0,+∞) Ask P what the heart of the triangle must be. O is a fixed point on the plane, A, B, C are three non-collinear points on the plane, and the moving point P satisfies OP=OA+λ(AB/|AB|cosB+AC/|AC|cosC).[0,+∞) | AB | cosB and | AC | cosC are the denominator of what the heart of a triangle must be. O is a fixed point on the plane, A, B, C are three non-collinear points on the plane, and the moving point P satisfies OP=OA+λ(AB/|AB|sinB+AC/|AC|sinC).[0,+∞) |AB|sinB and |AC|sinC are the denominator. Ask what center of the triangle P must pass through. The answers to the above three questions are the center of gravity It is required to provide the following other variants and expressions similar to the above with respect to the outer center and the outer center

Of your three results. The heart, the heart, the center of gravity, yes, but trouble, attention
| AB| sinB=|AC| sinC=BC, you can use:
OP=OA (AB+AC)0
P-point track passing the outer center:
OP=OA+AB/2[ CA/(} CA|cosA)+CB/(|CB|cosB)][ The method is the same as the perpendicular center, the following [] is the height on AB,λ is the real number, the trajectory of P is the center perpendicular of AB, passing through the outer center.]
Side center of point P locus passing through A:
OP=OA (AB/|AB AC/|AC|).[0,+∞)[ Exactly the same as your first expression
The bisector of ∠A goes through the heart, and through the side of A.]