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if B+C =60' show that

sin(120-B)=sin(120-C)

$L . H . S . = \sin (120^{0} - B) = \sin (120^{0} - (60^{0} - C)) (G i v e n, B + C = 60^{0} \Rightarrow B = 60^{0} - C) = \sin (120^{0} - 60^{0} + C) = \sin (60^{0} + C) = \sin (180^{0} - (60^{0} + C)) (\sin (180^{0} - θ) = \sin θ) = \sin (120^{0} - C) = R . H . S . H e n c e p r o v e d .$

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