Question

Considerando sen10°=0,17 e sen80°=0,98, calcule: cos10°; cos80°; tg10° e tg80°

Answer 1

$\sin^2x+\cos^2x=1\\\sin^210\degree + \cos^210\degree = 1\\(0,17)^2+\cos^210\degree=1\\0,0289 + \cos^210\degree=1\\\cos^210\degree = 1 - 0,0289\\\cos10\degree= \sqrt{0,9711}\\\boxed{\cos10\degree \approx 0,985}\\\\\sin^280\degree + \cos^280\degree = 1\\(0,98)^2+\cos^280\degree = 1\\0,9604 + \cos^280\degree = 1\\\cos^280\degree = 1 - 0,9604\\\cos80\degree = \sqrt{0,0396}\\\boxed{\cos80\degree \approx 0,198}$

$\tan x = \dfrac{\sin x}{\cos x}\\\\\\\tan10\degree = \dfrac{\sin 10 \degree}{\cos 10 \degree} = \dfrac{0,17}{0,985} \approx \boxed{0,1725}\\\\\\\tan80\degree = \dfrac{\sin 80\degree }{\cos 80\degree} = \dfrac{0,98}{0,198} \approx \boxed{4,949}$