Tam giác $ABC$ có đặc điểm gì, nếu các góc của nó thỏa  mãn hệ thức: $\frac{\sin C}{\sin B}=2 \cos A  ?$ 
Ta có: $\frac{\sin C}{\sin B}=2 \cos A\Leftrightarrow \sin C=2 \sin B \cos A$
Ta lại có $\sin (B+A)+\sin (B-A)$
$=\sin B \cos A+\sin A \cos B+\sin B \cos A-\sin A \cos B$
$=2 \sin B \cos A$
 ngoài ra, $B+A$ và $C$ bù nhau nên đẳng thức trên tương đương với
 $\sin C=\sin (B+A)+\sin (B-A)=\sin C+\sin (B-A)$
 $\Leftrightarrow \sin (B-A)=0\Leftrightarrow A=B$
 Tam giác $ABC$ có hai góc $A, B$ bằng nhau  nên là tam giác cân.
sinC/sinB=2cosA =>sinC=2cosA*sinB
=>sin(A-B)=0 =>A=B
=>ABC can tai C

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