Question

Calcule as medidas dos ângulos internos do vértices A(1,2) B (2,0) C (0,-1).

Answer 1

Temos os três vértices conhecidos do triângulo:

A(1, 2),  B(2, 0),  C(0, – 1).


Vamos encontrar as medidas a, b, c dos três lados do triângulo:

•   $\mathtt{c=d_{AB}}$

$\mathtt{c=d_{AB}}\\\\ \mathtt{c=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}}\\\\ \mathtt{c=\sqrt{(2-1)^2+(0-2)^2}}\\\\ \mathtt{c=\sqrt{1^2+(-2)^2}}\\\\ \mathtt{c=\sqrt{1+4}}\\\\ \mathtt{c=\sqrt{5}~u.c.}$


•   $\mathtt{b=d_{AC}}$

$\mathtt{b=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}}\\\\ \mathtt{b=\sqrt{(0-1)^2+(-1-2)^2}}\\\\ \mathtt{b=\sqrt{(-1)^2+(-3)^2}}\\\\ \mathtt{b=\sqrt{1+9}}\\\\ \mathtt{b=\sqrt{10}~u.c.}$


•   $\mathtt{a=d_{BC}}$

$\mathtt{a=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}}\\\\ \mathtt{a=\sqrt{(0-2)^2+(-1-0)^2}}\\\\ \mathtt{a=\sqrt{(-2)^2+(-1)^2}}\\\\ \mathtt{a=\sqrt{4+1}}\\\\ \mathtt{a=\sqrt{5}~u.c.}$

________

Pela Lei dos Cossenos,


•   Sendo $\mathtt{\alpha}$ o ângulo entre os lados que medem $\mathtt{b}$ e $\mathtt{c}:$

$\mathtt{a^2=b^2+c^2-2bc~cos\,\alpha}\\\\ \mathtt{cos\,\alpha=\dfrac{b^2+c^2-a^2}{2bc}}\\\\\\ \mathtt{cos\,\alpha=\dfrac{(\sqrt{10})^2+(\sqrt{5})^2-(\sqrt{5})^2}{2\cdot \sqrt{10}\cdot \sqrt{5}}}\\\\\\ \mathtt{cos\,\alpha=\dfrac{10+5-5}{2\sqrt{10\cdot 5}}}\\\\\\ \mathtt{cos\,\alpha=\dfrac{10}{2\sqrt{50}}}$

$\mathtt{cos\,\alpha=\dfrac{5}{5\sqrt{2}}}\\\\\\ \mathtt{cos\,\alpha=\dfrac{1}{\sqrt{2}}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathtt{\alpha=45^\circ}\end{array}}$


•   Sendo $\mathtt{\beta}$ o ângulo entre os lados que medem $\mathtt{a}$ e $\mathtt{c}:$

$\mathtt{b^2=a^2+c^2-2ac~cos\,\beta}\\\\ \mathtt{cos\,\beta=\dfrac{a^2+c^2-b^2}{2ac}}\\\\\\ \mathtt{cos\,\beta=\dfrac{(\sqrt{5})^2+(\sqrt{5})^2-(\sqrt{10})^2}{2\cdot \sqrt{5}\cdot \sqrt{5}}}\\\\\\ \mathtt{cos\,\alpha=\dfrac{5+5-10}{2\sqrt{5\cdot 5}}}\\\\\\ \mathtt{cos\,\beta=0}$

$\therefore~~\boxed{\begin{array}{c}\mathtt{\beta=90^\circ}\end{array}}$


•   Sendo $\mathtt{\gamma}$ o ângulo entre os lados que medem $\mathtt{b}$ e $\mathtt{c}:$

$\mathtt{c^2=a^2+b^2-2ab~cos\,\beta}\\\\ \mathtt{cos\,\gamma=\dfrac{a^2+b^2-c^2}{2ab}}\\\\\\ \mathtt{cos\,\gamma=\dfrac{(\sqrt{5})^2+(\sqrt{10})^2-(\sqrt{5})^2}{2\cdot \sqrt{5}\cdot \sqrt{5}}}\\\\\\ \mathtt{cos\,\gamma=\dfrac{5+10-5}{2\sqrt{5\cdot 10}}}\\\\\\ \mathtt{cos\,\gamma=\dfrac{10}{2\sqrt{50}}}$

$\mathtt{cos\,\gamma=\dfrac{1}{\sqrt{2}}}\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathtt{\gamma=45^\circ}\end{array}}$


________

Eis os ângulos internos do triângulo:

$\mathtt{\alpha=45^\circ,~\beta=90^\circ,~\gamma=45^\circ.}$


Dúvidas? Comente.


Bons estudos! :-)

Answer 2

$Vou\hspace{3}chamar\hspace{3}os\hspace{3}\^angulos\hspace{3}de\hspace{3}alpha(\alpha),\hspace{3}beta(\beta)\hspace{3}e\hspace{3}gamma(\gamma).\hspace{3}Irei\hspace{3}de-\\terminar\hspace{3}o\hspace{3}\^angulo\hspace{3}\alpha\hspace{3}como\hspace{3}o\hspace{3}\^angulo\hspace{3}entre\hspace{3}os\hspace{3}segmentos\hspace{3}\overrightarrow{AB}\hspace{3}e\hspace{3}\overrightarrow{AC}.$

$\overrightarrow{AB}=B-A=2-1,0-2=(1,-2) \\ \overrightarrow{AC}=C-A=0-1,-1-2=(-1,-3)\\\\cos\hspace{3}\alpha= \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|\overrightarrow{AB}|\cdot |\overrightarrow{AC}|} = \frac{(1,-2)\cdot(-1, -3)}{ \sqrt{1^2+(-2)^2}\cdot \sqrt{(-1)^2+(-3)^2}}= \frac{-1+6}{\sqrt{5}\cdot \sqrt{10}}= \frac{5}{ \sqrt{50}}\\\\= \frac{5}{ \sqrt{50}}\cdot \frac{\sqrt{50}}{ \sqrt{50}}= \frac{5\sqrt{50}}{50}=\frac{ 5\sqrt{2\cdot 25}}{50}=\frac{5\cdot5 \sqrt{2}}{50}=\frac{25 \sqrt{2}}{50}= \frac{\sqrt{2}}{2},\hspace{3}logo:$ $cos\hspace{3}\alpha= \frac{\sqrt{2}}{2},\hspace{1}o\hspace{3}\^angulo\hspace{3}que\hspace{3}possui\hspace{3}esse\hspace{3}cosseno\hspace{3}\'e\hspace{3}um\hspace{3}\^angulo\hspace{3}not\'avel\\\\ de\hspace{3}\hspace{3}45^\circ .\hspace{5}Portanto\hspace{3}\alpha=45^\circ$

$Agora\hspace{3}irei\hspace{3}determinar\hspace{3}\beta\hspace{3}como\hspace{3}um\hspace{3}\^angulo\hspace{3}entre\hspace{3}os\hspace{3}segmentos\\ \overrightarrow{BA}\hspace{3}e\hspace{3}\overrightarrow{BC}:\\\\ \overrightarrow{BA}=A-B=1-2,\hspace{3}2-0=(-1,2)\\ \overrightarrow{BC}=C-B=0-2,\hspace{3}-1-0=(-2,-1)\\\\ cos\hspace{3} \beta= \frac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{|\overrightarrow{BA}|\cdot|\overrightarrow{BC}|}=\frac{(-1,2)\cdot (-2,-1)}{ \sqrt{(-1)^2+2^2}\cdot \sqrt{(-2)^2+(-1)^2}}= \frac{2-2}{\sqrt{5}\cdot \sqrt{5}}= 0$$O\hspace{3}\^angulo\hspace{3}que\hspace{3}possui\hspace{3}cosseno=0,\hspace{3}\'e\hspace{3}o\hspace{3}\^angulo\hspace{3}not\'avel\hspace{3}de\hspace{3}90^\circ;\\portanto\hspace{3}o\hspace{3}\^angulo\hspace{3}\beta =90^\circ.\\\\Se\hspace{3}\alpha=45^\circ,\hspace{3}e\hspace{3}\beta=90^\circ,\hspace{3}o\hspace{3}\^angulo\hspace{3}\gamma\hspace{3}s\'o\hspace{3}pode$$\hspace{3}ser\hspace{3}de\hspace{3}45^\circ,\hspace{3}pois\hspace{3}\alpha +\beta +\gamma=180^\circ\hspace{3}$$Concluindo,\hspace{3}as\hspace{3}medidas\hspace{3}dos\hspace{3}\^angulos\hspace{3}internos\hspace{3}do\hspace{3}tri\^angulo\\de\hspace{3}v\'ertices\hspace{3}A(1,2),\hspace{3}B(2,0)\hspace{3}e\hspace{3}C(0,-1)\hspace{3}s\~ao:\\\\ \alpha=45^\circ,\hspace{3}\beta=90^\circ\hspace{3}e\hspace{3}\gamma=45^\circ$