Question

в треугольнике ABC угол C равен 90 sinA= 5/√41. найти tg B

Answer 1

$sin A=cos B=frac{5}{sqrt{41}}; 0^o<B<90^o;<var></var></p> <p>sin B=sqrt{1-cos^2 B}=sqrt{1-(frac{5}{sqrt{41}})^2}=frac{4}{sqrt{41}};</p> <p>tg B=frac{sin B}{cos B}=frac{frac{4}{sqrt{41}}}{frac{5}{sqrt{41}}}=frac{4}{5}=0.8$