Question

sin(Пи-A)=0.6                    A принадлежит (90:180)
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tg2A

Answer 1

$sin(\pi - \alpha )=sin \alpha =0,6\\\\cos \alpha =-\sqrt{1-sin^2 \alpha }=-\sqrt{1-0,36}=-0,8\; ,\;\; t.k.\; 90\ \textless \ \alpha \ \textless \ 180\\\\tg \alpha =-\frac{0,6}{0,8}=-0,75=-\frac{3}{4} \\\\tg2 \alpha =\frac{2tg \alpha }{1-tg^2 \alpha }=\frac{-\frac{6}{4}}{1-\frac{9}{16}}=\frac{-6\cdot 16}{7\cdot 4}=\frac{-24}{7}=-3,42857$

Answer 2

$sin( \pi - \alpha )=0.6 \\ sin( \pi - \alpha )=sin \alpha \\ sina=0.6 \\ \\ cosa=- \sqrt{1-sin ^{2} \alpha } =- \sqrt{1-0.6 ^{2} } =- \sqrt{1-0.36} = \\ =- \sqrt{0.64} =-0.8 \\ \\ \\ tg \alpha = \frac{sin \alpha }{cos \alpha } = \frac{0.6}{-0.8} =- \frac{0.6}{0.8} =- \frac{6}{8} =- \frac{3}{4}$

$\\ tg2 \alpha = \frac{2tg \alpha }{1-tg ^{2} \alpha } = \frac{2*(- \frac{3}{4}) }{1-( -\frac{3}{4} ) ^{2} } = \frac{- \frac{3}{2} }{1- \frac{9}{16} } =- \frac{ \frac{3}{2} }{ \frac{16}{16}- \frac{9}{16} } = \\ =- \frac{ \frac{3}{2} }{ \frac{7}{16} } =- \frac{3}{2} / \frac{7}{16} =- \frac{3*16}{2*7} =- \frac{3*8}{7} =- \frac{24}{7} =- 3\frac{3}{7}$