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2014-03-25T14:52:35+07:00
Kuadran III: sin = - , cos = - , tan = +

Segitiga ABC siku di B,
sin A =  \frac{BC}{AC} =  \frac{8}{10}
AB =  \sqrt{ AC^{2}- BC^{2}  }  =  \sqrt{ 10^{2} -  8^{2}  } = 6

cos A =  \frac{AB}{AC} =  \frac{6}{10}  = -0.6 (cos negatif di kuadran III)
tan A =  \frac{BC}{AB} =  \frac{8}{6} = 1.33 (tan positif di kuadran III)