(cosB/sinC)* vector AB+(cosC/sinB)* vector AC=2m* vector AO can be transformed into:

(cosB/sinC)* (vector OB- vector OA)+(cosC/sinB)* (vector OC-OA)=-2m* vector OA (*).

It is easy to know that the angle between vector OB and OA is 2∠C, that between vector OC and OA is 2∠B, and that between vector OA and OA is 0.

Vector OA|=| Vector ob| =| Vector OC | = R.

Then multiply the left and right sides of the formula (*) by the vector OA respectively, and you can get:

(cosB/sinC)* (vector OB* vector OA- vector OA* vector OA)+(cosC/sinB)* (vector OC* vector OA- vector OA* vector OA) =-2m * (vector OA* vector OA).

That is, (cosB/sinC)*R? (cos2C - 1)+(cosC/sinB)*R？ (cos2B - 1)=-2m*R？

2sin？ C*cosB/sinC +2sin？ B*cosC/sinB=2m

sinC*cosB+sinB*cosC=m

sin(B+C)=m

Because Sina = sin [π-(b+c)] = sin (b+c) and ∠A=θ.

So m=sinA=sinθ.

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