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Prove that

sin50° - sin70° + sin10° = 0

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Show that sin70°+sin50°= √3cos0°
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the triangle CDE, m<D=90 degrees, m<C=50 degrees, and ED=23. Which equation could be used to find CE?1. sin50=CE/23 2.
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[Cos70°- cos60°]÷[sin 70°- sin50°]= -√3
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