Question

Determine sen B,se cos B=0,6

Answer 1

Explicação passo-a-passo:

${sen}^{2} b + cos ^{2} b = 1 \\ {sen}^{2} b+ {0.6}^{2} = 1 \\ {sen}^{2} b + 0.36 = 1 \\ {sen}^{2} b= 1 - 0.36 \\ {sen}^{2}b = 0.64 \\ sen ^{2} b= \dfrac{64}{100} \\ senb = \sqrt{ \dfrac{64}{100} } \\ \boxed{ \boxed{senb =\sqrt{ \dfrac{64}{100} }} = > \frac{8}{10} = > \frac{4}{5} }$