Question

Для острого угла a найдите sin a,cos oc и ctgoc если tg oc=1/4. помогите​

Answer 1

Объяснение:

$tg\alpha =\frac{1}{4}\ \ \ \ 0^0<\alpha 90^0\ \ \ \ sin\alpha =?\ \ \ \ cos\alpha =?\ \ \ \ ctg\alpha=?\\ctg\alpha =\frac{1}{tg\alpha } =\frac{1}{\frac{1}{4} } =4.\\(tg^2\alpha )=(\frac{1}{4})^2\\\frac{sin^2\alpha }{cos^2\alpha } =\frac{1}{16} \\16*sin^2\alpha =cos^2\alpha \\16*sin^2\alpha+sin^2\alpha =cos^2\alpha+sin^2\alpha \\17*sin^2\alpha =1\ |:17\\sin^2\alpha =\frac{1}{17}\\sin\alpha =б\sqrt{\frac{1}{17} }=б\frac{1}{\sqrt{17} }=б\frac{\sqrt{17} }{17}$

$sin\alpha =\frac{\sqrt{17} }{17} \ \ \ \ (0^0<\alpha <90^0)\\cos^2\alpha=1-sin^2\alpha \\cos^2\alpha =1-(\frac{\sqrt{17} }{17})^2=1-\frac{1}{17} =\frac{16}{17}\\cos\alpha =б\sqrt{\frac{16}{17} } =б\frac{4}{\sqrt{17} } =б\frac{4\sqrt{17} }{17}. \\cos\alpha =\frac{4\sqrt{17} }{17}.\ \ \ \ (0^0<\alpha <90^0).$

Ответ: $sin\alpha =\frac{\sqrt{17} }{17},\ \ cos\alpha =\frac{4\sqrt{17} }{17} ,\ \ ctg\alpha =4.$