Question

Sabendo que sem 20 = 0,34, calcule:
a) tg 20°
b) sen 40°

Answer 1

sen(20º) = 0,34 = 34/100 = 17/50

a)  $sen^2(20^o) + cos^2(20^o) = 1 \\ \\ (\frac{17}{50})^5 + cos^2(20^o) = 1 \\ \\ cos^2(20^o) = 1 - \frac{289}{2500} \\ \\ cos(20^o) = \sqrt{\frac{2211}{2500} } ==> \frac{\sqrt{2211}}{50} \\ \\ tg(20^o) = \frac{ \frac{17}{50} }{ \frac{\sqrt{2211}}{50} } \\ \\ tg(20^o) = \frac{17}{50}.\frac{50}{\sqrt{2211}} ==> \frac{17}{\sqrt{2211}}$

b)

$sen(40^o) = sen(2.20^o) = 2.sen(20^o).cos(20^o) \\ \\ sen(40^o) = 2.\frac{17}{50}.\frac{\sqrt{2211}}{50} ===> sen(40^o) = \frac{17\sqrt{2211}}{1250}$