Question

Gostaria da resolução, abraço!

Answer 1

$\alpha+\beta+\theta=180^\circ$


Como $\alpha,$ $\beta$ e $\theta$ são diretamente proporcionais a $1,$ $2$ e $3,$ devemos ter

$\bullet\;\;\alpha=\dfrac{1}{1+2+3}\cdot 180^\circ\\\\\\ \alpha=\dfrac{1}{6}\cdot 180^\circ\\\\\\ \boxed{\begin{array}{c}\alpha=30^\circ \end{array}}$


$\bullet\;\;\beta=\dfrac{2}{1+2+3}\cdot 180^\circ\\\\\\ \beta=\dfrac{2}{6}\cdot 180^\circ\\\\\\ \boxed{\begin{array}{c}\beta=60^\circ \end{array}}$


$\bullet\;\;\theta=\dfrac{3}{1+2+3}\cdot 180^\circ\\\\\\ \theta=\dfrac{3}{6}\cdot 180^\circ\\\\\\ \boxed{\begin{array}{c}\theta=90^\circ \end{array}}$


Então, temos um triângulo retângulo, cujos ângulos internos são

$\left\{ \!\begin{array}{l}\alpha=30^\circ\\\beta=60^\circ\\\theta=90^\circ \end{array} \right.$

______________

Ver figura em anexo.

O segmento $BP$ é a bissetriz do ângulo $\beta=60^\circ.$

Por razões trigonométricas, encontramos as medidas dos outros segmentos:

$\bullet\;\;BC=BP\cdot \mathrm{sen\,}60^\circ\\\\ BC=2\cdot \dfrac{\sqrt{3}}{2}\\\\\\ \boxed{\begin{array}{c}BC=\sqrt{3}\mathrm{~u.c.} \end{array}}$


$\bullet\;\;\dfrac{BC}{AB}=\mathrm{sen\,}30^\circ\\\\\\ AB=\dfrac{BC}{\mathrm{sen\,}30^{\circ}}\\\\\\ AB=\dfrac{\sqrt{3}}{\frac{1}{2}}\\\\\\ \boxed{\begin{array}{c}AB=2\sqrt{3}\mathrm{~u.c.} \end{array}}$


$\bullet\;\;\dfrac{BC}{AC}=\mathrm{tg\,}30^\circ\\\\\\ AC=\dfrac{BC}{\mathrm{tg\,}30^\circ}\\\\\\ AC=\dfrac{\sqrt{3}}{\frac{\sqrt{3}}{3}}\\\\\\ AC=\sqrt{3}\cdot \dfrac{3}{\sqrt{3}}\\\\\\ \boxed{\begin{array}{c}AC=3\mathrm{~u.c.} \end{array}}$

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O perímetro do triângulo é

$AB+BC+AC\\\\ =2\sqrt{3}+\sqrt{3}+3\\\\ =3\sqrt{3}+3\\\\ =\boxed{\begin{array}{c}3(\sqrt{3}+1)\mathrm{~u.c.} \end{array}}$


Bons estudos! :-)

Answer 2

A solução está no anexo.