Question

Задание на картинке ⬆️⬆️⬆️ Номер 204

Answer 1

$\dfrac{\pi }{6}<\alpha<\dfrac{7\pi }{6}\\\\\alpha-\dfrac{\pi }{6} =x\\\\Cosx=\dfrac{3}{5}\\\\Sinx=\sqrt{1-Cos^{2}x } =\sqrt{1-\Big(\dfrac{3}{5}\Big)^{2}}=\sqrt{1-\dfrac{9}{25} } =\\\\=\sqrt{\dfrac{16}{25} }=\dfrac{4}{5}\\\\\alpha=\dfrac{\pi }{6} +x\\\\Sin\alpha=Sin\Big( \dfrac{\pi }{6} -x\Big)=Sin\dfrac{\pi }{6}Cosx-Sinx Cos\dfrac{\pi }{6}=\dfrac{1}{2}\cdot\dfrac{3}{5}-\dfrac{4}{5}\cdot\dfrac{\sqrt{3} }{2} =\\\\=\dfrac{3}{10}-\dfrac{4\sqrt{3} }{10}=\boxed{\dfrac{3-4\sqrt{3} }{10}}$

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