If a, b, c are the sides of a ∆ ABC opposite angles A,B,C respectively, and Δ$$=a2bsin$$⁡Acsin⁡Absin⁡$$A1cos$$⁡$$(B$$−$$C)csin$$⁡Acos⁡$$(B$$−$$C)1$$, then ∆equals

Question

If a, b, c are the sides of a ∆ ABC opposite angles A,B,C respectively, and Δ$$=a2bsin$$⁡Acsin⁡Absin⁡$$A1cos$$⁡$$(B$$−$$C)csin$$⁡Acos⁡$$(B$$−$$C)1$$, then ∆equals

Infinity Learn · Accepted Answer

By the law of sinesasin⁡$$A=bsin$$⁡$$B=csin$$⁡$$C=k(say)$$⇒$$a=ksin$$⁡A,$$b=ksin$$⁡B,$$c=ksin$$⁡C. Now Δ$$=a2ab/kac/kab/k1cos$$⁡(B$$−$$C)ac/kcos$$⁡$$(B$$−$$C)1=a21sin$$⁡Bsin⁡Csin⁡$$B1cos$$⁡$$(B$$−$$C)$$sin$$⁡Ccos⁡(B$$−$$C)1=a21sin$$⁡$$(A+C)$$sin$$⁡(A+B)$$sin$$⁡(A+C)1cos$$⁡$$(B$$−$$C)$$sin$$⁡(A+B)$$cos$$⁡$$(B$$−$$C)1=a2sin$$⁡Acos⁡$$A0cos$$⁡Csin⁡$$C0cos$$⁡Bsin⁡$$B0sin$$⁡Acos⁡$$A0cos$$⁡Csin⁡$$C0cos$$⁡Bsin⁡$$B0=a2(0)=0$$

If $a, b, c$ are the sides of a $∆ A B C$ opposite angles $A, B, C$ respectively, and $Δ = |\begin{array}{ccc} a^{2} & b \sin A & c \sin A \\ b \sin A & 1 & \cos (B - C) \\ c \sin A & \cos (B - C) & 1 \end{array}|,$ then $∆$ equals

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If a, b, c are the sides of a ∆ ABC opposite angles A,B,C respectively, and Δ=a2bsin⁡Acsin⁡Absin⁡A1cos⁡(B−C)csin⁡Acos⁡(B−C)1, then ∆equals

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If $a, b, c$ are the sides of a $∆ A B C$ opposite angles $A, B, C$ respectively, and $Δ = |\begin{array}{ccc} a^{2} & b \sin A & c \sin A \\ b \sin A & 1 & \cos (B - C) \\ c \sin A & \cos (B - C) & 1 \end{array}|,$ then $∆$ equals