Question

How do I solve this

Answer 1

`sine=(opposite)/(hypotenuse)\\\\tangent=(opposite)/(adjacent)`

`\sin B=(b)/(a);\ \tan C=(c)/(b)\\\\\sin B\cdot\tan C=(b)/(a)\cdot(c)/(b)=(c)/(a)\\\\\sin C=(c)/(a);\ \tan B=(b)/(c)\\\\\sin C\cdot\tan B=(c)/(a)\cdot(b)/(c)=(b)/(a)`