Find the known trilateration angle formula cosA b 2 c 2 a 2 2bc derivation steps

Here's the answer:Solution: Polar coordinates.

It is to take a fixed point o (pole) in the plane to lead a pure ray of missing ray ox (polar axis), and then select a unit of length.

and the positive direction of the angle (usually taken counterclockwise), a description of the ordinal number pair (. Represents the distance, 0. In this problem, =2a(2+cos) 0.

Under the premise that a is a positive value, 2+cos has a minimum positive period in its slip.

0,2], therefore, the integral interval is [0,2]. In the same way, =2a(1+cos) 0, satisfying 0 is [0, 2] [3 2,2 ], and the interval of the argument is [- 2, 2] for reference.

The title should be in the triangle, a, b, c, are the three burning numbers of the triangle. By the formula a sina = b sinb sin b sinc

2b-c)cosa-acosc=0 can be reduced to biyin 2sinbcosa-sinccosa-sinacosc=0

2sinbcosa=sinccosa+sinacosc2sinbcosa=sin（a+c）

Because in the triangular skin containment, a+b+c=180 degrees, a+c=180-b has sin(a+c)=sinb

i.e. 2sinbcosa = sinb

2cosa=1

cosa=1/2

a=60 degrees.

cos(180+a)=-cosa

cos(-a)=cosa

cos(2a) = cosa 2-sina 2

cos(180-2a)=-cos(-2a)=-cos(2a)=-cos a)2 - sin a)2] =sin a)2 - cos a)2

Let (cos a)2=c, then, (sin a)2=1-c, then the original formula is: c+( 1-c-c) 2 =1c+1-4c+4c 2 =1

4c^2 - 3c +1 =1

4c^2 - 3c =0

c(4c-3)=0

c = 0 or hermit c = 3 4

Namely. cos a=0 or cos a = root3 2 or cos a = negative root3 2

A=90 or 270 can be obtained

or a=30 or 330

or a=150 or 210

i.e.: a=k + 6

or town belt. a=k∏-∏6

Or. a=k∏+∏2

1+cos(a)]/2

1/2+1/2cos(a/2+a/2)

1 2+1 2[cos 2(a defeat or 2)-sin 2(a 2)]1 2-1 2sin 2(a chavu 2)+1 2cos 2(a 2)1 2cos 2(a 2)+1 2cos 2(a 2) cos 2(a 2)cos 2(a 2)

I'm exhausted.

0°b (big angle to big side)

sina/a=sinb/b

sinb=1 old stuffy2

b=30°c=180-(a+b)=105°

If it is cosa=2 3

A=arccos(2 Pulsator 3).

or cos(-a)=2 basis rise or defeat 3

a= -arccos(2 laughs 3).

a=2kπ±arccos(2/3)