Question

1- Sabendo que Sen 20°= 0,345 calcule. Sen, Cos e Tg de 70°. (Use essas informações no próximo exercício).

2- Responda as questões:
a) Um avião levanta e vai formando um ângulo de 20° com o solo. Qual a sua altura após percorrer 1000 m na horizontal? e após 1500 m
b) O piloto deseja atingir 700 m após 2000 m
no horizontal. Calcule sen, cos e tg do ângulo formado pelo avião. ​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf Seno,cosseno\,e\,tangente}\\\underline{\sf de\,\hat angulos\,complementares}\\\sf dois\,\hat angulos\,\theta\,e\,\alpha\,s\tilde ao\,complementares\\\sf quando\,\theta+\alpha=90^\circ\\\sf observe\,que~20^\circ+70^\circ=90^\circ\\\sf sobre\,os\,\hat angulos\,complementares\,vale\,as\,identidades:\\\sf\bullet ~sen(90^\circ-\theta)=cos(\theta)\\\sf\bullet~cos(90^\circ-\theta)=sen(\theta)\\\sf \bullet ~tg(90^\circ-\theta)=\dfrac{1}{tg(\theta)}\end{array}}$

$\large\boxed{\begin{array}{l}\rm 1)~\sf Vou\,admitir\,que\,estes\,\hat angulos\\\sf pertenc_{\!\!,}am\,ao\,tri\hat angulo\,ret\hat angulo.\\\sf sen^2(20^\circ)+cos^2(20^\circ)=1\\\sf cos(20^\circ)=\sqrt{1-sen^2(20^\circ)}\\\sf cos(20^\circ)=\sqrt{1-0,345^2}\\\sf cos(20^\circ)=\sqrt{1-0,119025}\\\sf cos(20^\circ)=\sqrt{0,880975}\\\sf cos(20^\circ)=0,938\\\sf tg(20^\circ)=\dfrac{sen(20^\circ)}{cos(20^\circ)}=\dfrac{0,345}{0,938}=0,367\end{array}}$

$\large\boxed{\begin{array}{l}\sf sen(90^\circ-20^\circ)=cos(20^\circ)\\\sf sen(70^\circ)=0,938\\\sf cos(90^\circ-20^\circ)=sen(20^\circ)\\\sf cos(70^\circ)=0,345\\\sf tg(90^\circ-20^\circ)=\dfrac{1}{tg(20^\circ)}\\\sf tg(70^\circ)=\dfrac{1}{0,367}\\\\\sf tg(70^\circ)=2,724\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2)~\sf perceba\,que\,a\,altura\,\acute e~o~cateto~oposto\,a\,20^\circ\\\sf e~1000\,m~o~cateto~adjacente~a~20^\circ.\\\sf podemos\,usar\,a\,relac_{\!\!,}\tilde ao\,tangente\\\sf tg(20^\circ)=\dfrac{h}{1000}\implies h=1000\cdot tg(20^\circ)\\\\\sf h=1000\cdot0,367\\\sf h=367\,m\end{array}}$

$\large\boxed{\begin{array}{l}\sf Quando\,ele\,percorrer\,1500\,m,teremos\,a\,mesma\\\sf situac_{\!\!,}\tilde ao\,que\,a\,anterior,podemos\,usar\\\sf a\,tangente\,para\,calcular\,a\,altura.\\\sf tg(20^\circ)=\dfrac{h}{1500}\implies h=1500 tg(20^\circ)\\\\\sf h=1500\cdot0,367\\\sf h=550,5~m\end{array}}$

$\large\boxed{\begin{array}{l}\sf Aqui\,ele\,j\acute a\,percorreu\,2000\,m\,na\,horizontal\\\sf e\,deseja\,atingir\, 700\,m\,de\,altura\\\sf e\,quer\,saber\,qual\,deve\,ser\,o\\\sf seno,cosseno\,e\,tangente\,do\,\hat angulo\\\sf para\,que\,isso\,ocorra.\\\sf calculando\,a\,hipotenusa\,pelo\,teorema\,de\,Pit\acute agoras\,temos: \end{array}}$

$\large\boxed{\begin{array}{l}\sf x^2=2000^2+700^2\\\sf x^2=4~000~000+490~000\\\sf x^2=4490000\\\sf x=\sqrt{4490000}\\\sf x=2118,96\approxeq2119\,m\\\sf chamando\,o\,\hat angulo\,de\,\theta\,temos:\\\sf sen(\theta)=\dfrac{700}{2219}\\\\\sf cos(\theta)=\dfrac{2000}{2119}\\\\\sf tg (\theta)=\dfrac{700}{2000}=\dfrac{7}{20}\end{array}}$