Let \(\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}\), where \(A, B, C\) are angles of a t....
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Let \(\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}\), where \(A, B, C\) are angles of a triangle \(A B C\). If the lengths of the sides opposite these angles are \(a, b, c\) respectively, then: 📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live
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