Question

calcule as medidas a e b nos triângulos retângulos​

Answer 1

$\Large\boxed{\begin{array}{l}\tt a)~\sf tg(45^\circ)=\dfrac{a}{8}\\\sf 1=\dfrac{a}{8}\\\sf a=8\cdot1=8\\\sf cos(45^\circ)=\dfrac{8}{b}\\\sf\dfrac{\sqrt{2}}{2}=\dfrac{8}{b}\\\\\sf \sqrt{2}b=16\\\\\sf b=\dfrac{16}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf b=\dfrac{\diagdown\!\!\!\!\!\!16\sqrt{2}}{\backslash\!\!\!2}\\\\\sf b=8\sqrt{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf tg( 60^\circ)=\dfrac{6\sqrt{3}}{a}\\\\\sf \diagup\!\!\!\!\!\!\sqrt{3}=\dfrac{6~\diagup\!\!\!\!\!\!\sqrt{3}}{a}\\\\\sf a=6\cdot1=6\\\\\sf cos(60^\circ)=\dfrac{a}{b}\\\\\sf \dfrac{1}{2}=\dfrac{6}{b}\\\\\sf b=2\cdot6\\\\\sf b=12\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)~\sf sen(30^\circ)=\dfrac{12}{b}\\\\\sf \dfrac{1}{2}=\dfrac{12}{b}\\\\\sf b=2\cdot12\\\sf b=24\\\\\sf cos(30^\circ)=\dfrac{a}{b}\\\\\sf \dfrac{\sqrt{3}}{2}=\dfrac{a}{24}\\\\\sf 2a=24\sqrt{3}\\\\\sf a=\dfrac{24\sqrt{3}}{2}\\\\\sf a=12\sqrt{3}\end{array}}$