Question

3.Для острого угла α найдите sinα, cosα, tgα, если ctg α= 1/3

чертеж нужен тоже ​

Answer 1

Объяснение:

$ctg\alpha =\frac{1}{3} \ \ \ \ 0^0<\alpha<90^0\ \ \ \ sin\alpha =?\ \ \ \ cos\alpha =?\ \ \ \ tg\alpha =?\\tg\alpha =\frac{1}{ctg\alpha }=\frac{1}{\frac{1}{3} }=3.\\tg^2\alpha =(\frac{sin\alpha }{cos\alpha } )^2=\frac{sin^2\alpha }{cos^2\alpha } =3^2=9.\\\frac{sin^2\alpha }{cos^2\alpha }=9\\sin^2\alpha =9*cos^2\alpha \\ sin^2\alpha+cos^2\alpha =9*cos^2\alpha +cos^2\alpha\\1=10*cos^2\alpha \ |:10\\cos^2\alpha =\frac{1}{10}.$

$cos\alpha =б\sqrt{\frac{1}{10} }=б{\frac{1}{\sqrt{10} } } =б\frac{\sqrt{10} }{10}.\\cos\alpha =\frac{\sqrt{10} }{10}\ \ \ \ (0^0<\alpha <90^0).\\sin^2\alpha =1-cos^2\alpha =1-\frac{1}{10}=\frac{9}{10} \\sin\alpha =б\sqrt{\frac{9}{10} } =б\frac{3}{\sqrt{10} } =б\frac{3\sqrt{10} }{10}. \\sin\alpha =\frac{3\sqrt{10} }{10} \ \ \ \ (0^0<\alpha <90^0).$

Ответ: sinα=3√10/10, cosα=√10/10, tgα=3.