Question

Usando o sen/cos/tg descubra o valor da incógnita em cada figura.​

Answer 1

$\boxed{\begin{array}{l}\tt 1)~\sf sen(30^\circ) =\dfrac{x}{120}\\\sf\dfrac{1}{2}=\dfrac{x}{120}\\\sf x=\dfrac{120}{2}\\\sf x=60~m\\\tt 2)~\sf tg(30^\circ)=\dfrac{x}{30}\\\sf\dfrac{\sqrt{3}} {3}=\dfrac{x}{30}\\\sf x=\dfrac{30\sqrt{3}}{3}\\\sf x=10\sqrt{3}\\\tt 3)~\sf sen(45^\circ)=\dfrac{x}{50}\\\sf\dfrac{\sqrt{2}}{2}=\dfrac{x}{50}\\\sf x=\dfrac{50\sqrt{2}}{2}\\\sf x=25\sqrt{2}~cm\\\tt 4)~\sf tg(30^\circ)=\dfrac{\sqrt{3}}{y} \\\sf\dfrac{\diagup\!\!\!\!\sqrt{3}}{3}=\dfrac{\diagup\!\!\!\!\sqrt{3}}{y}\\\sf y=3\end{array}}$

$\boxed{\begin{array}{l}\tt 5)~\sf tg(30^\circ) =\dfrac{x}{9}\\\sf\dfrac{\sqrt{3}}{3}=\dfrac{x}{9}\\\sf x=\dfrac{9\sqrt{3}}{3}\\\sf x=3\sqrt{3}\\\tt 6)~\sf sen(60^\circ)=\dfrac{12\sqrt{3}}{x}\\\sf\dfrac{\diagup\!\!\!\!\sqrt{3}}{2}=\dfrac{12\diagup\!\!\!\!\sqrt{3}}{x}\\\sf x=2\cdot12\\\sf x=24\\\tt 7)~\sf x^2=14^2+48^2\\\sf x^2=196+2304\\\sf x^2=2500\\\sf x=\sqrt{2500}\\\sf x=50\\\tt 8)~\sf x^2=(3\sqrt{2}) ^2+(3\sqrt{2})^2\\\sf x^2=18+18\\\sf x^2=36\\\sf x=\sqrt{36}\\\sf x=6\end{array}}$

$\boxed{\begin{array}{l}\tt 9)~\sf y^2+2^2=16^2\\\sf y^2+4=256\\\sf y^2=256-4\\\sf y^2=252\\\sf y=\sqrt{252}\\\sf y=6\sqrt{7}\\\tt 10)~\sf cos(45^\circ)=\dfrac{x}{20\sqrt{2}}\\\sf \dfrac{\sqrt{2}}{2}=\dfrac{x}{20\sqrt{2}}\\\sf x=\dfrac{20\sqrt{2}\cdot\sqrt{2}}{2}\\\sf x=\dfrac{20\cdot\diagup\!\!\!2}{\diagup\!\!\!2}\\\sf x=20\\\sf y=x=20\\\tt 11)~\sf cos(30^\circ) =\dfrac{9\sqrt{3}}{y}\\\sf\dfrac{\diagup\!\!\!\!\sqrt{3}}{2}=\dfrac{9\diagup\!\!\!\!\sqrt{3}}{y}\\\sf y=2\cdot9\\\sf y=18\\\tt 12)~\sf sen(30^\circ)=\dfrac{x}{1000}\\\sf\dfrac{1}{2}=\dfrac{x}{1000}\\\sf x=\dfrac{1000}{2}\\\sf x=500~m \end{array}}$