Question

No interior de São Paulo, dois observadores A e B visualizam um balão sob ângulos de 30° e 45°. Sabendo que a distância entre eles é de 100 metros, qual é a altura do balão em questão?:

Answer 1

Observe a figura em anexo:

O balão está posicionado no ponto $C,$ e a altura $h$ procurada é o comprimento do segmento $CM.$


$\bullet\;\;$ Do triângulo $ACM,$ tiramos que

$\mathrm{tg\,}30^\circ=\dfrac{CM}{AM}\\\\\\ \mathrm{tg\,}30^\circ=\dfrac{h}{100-x}\\\\\\ (\mathrm{tg\,}30^\circ)\cdot (100-x)=h\\\\ 100\,\mathrm{tg\,}30^\circ-x\,\mathrm{tg\,}30^\circ=h\\\\ 100\,\mathrm{tg\,}30^\circ-h=x\,\mathrm{tg\,}30^\circ\\\\ x=\dfrac{100\,\mathrm{tg\,}30^\circ-h}{\mathrm{tg\,}30^\circ}~~~~~~\mathbf{(i)}$


$\bullet\;\;$ Do triângulo $BCM,$ tiramos que

$\mathrm{tg\,}45^\circ=\dfrac{CM}{BM}\\\\\\ \mathrm{tg\,}45^\circ=\dfrac{h}{x}\\\\\\ x\,\mathrm{tg\,}45^\circ=h\\\\ x=\dfrac{h}{\mathrm{tg\,}45^\circ}~~~~~~\mathbf{(ii)}$

____________________

Igualando $\mathbf{(i)}$ e $\mathbf{(ii)},$ devemos ter

( resolvendo a equação para $h$ )

$\dfrac{100\,\mathrm{tg\,}30^\circ-h}{\mathrm{tg\,}30^\circ}=\dfrac{h}{\mathrm{tg\,}45^\circ}\\\\\\ h\,\mathrm{tg\,}30^\circ=\mathrm{tg\,}45^\circ\cdot (100\,\mathrm{tg\,}30^\circ-h)\\\\ h\,\mathrm{tg\,}30^\circ=100\,\mathrm{tg\,}30^\circ\,\mathrm{tg\,}45^\circ-h\,\mathrm{tg\,}45^\circ\\\\ h\,\mathrm{tg\,}30^\circ+h\,\mathrm{tg\,}45^\circ=100\,\mathrm{tg\,}30^\circ\,\mathrm{tg\,}45^\circ\\\\ h\cdot (\mathrm{tg\,}30^\circ+\,\mathrm{tg\,}45^\circ)=100\,\mathrm{tg\,}30^\circ\,\mathrm{tg\,}45^\circ\\\\ h=\dfrac{100\,\mathrm{tg\,}30^\circ\,\mathrm{tg\,}45^\circ}{\mathrm{tg\,}30^\circ+\,\mathrm{tg\,}45^\circ}$


Substituindo os valores de $\mathrm{tg\,}30^\circ$ e $\mathrm{tg\,}45^\circ,$ temos

$h=\dfrac{100\cdot \frac{\sqrt{3}}{3}\cdot 1}{\frac{\sqrt{3}}{3}+1}\\\\\\ h=\dfrac{\frac{100\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}+1}\cdot \dfrac{3}{3}\\\\\\ h=\dfrac{100\sqrt{3}}{\sqrt{3}+3}\\\\\\ h=\dfrac{100\sqrt{3}}{\sqrt{3}+\sqrt{3}\cdot \sqrt{3}}\\\\\\ h=\dfrac{100\sqrt{3}}{\sqrt{3}\cdot (1+\sqrt{3})}\\\\\\ h=\dfrac{100}{1+\sqrt{3}}\\\\\\ h=\dfrac{100}{1+\sqrt{3}}\cdot \dfrac{1-\sqrt{3}}{1-\sqrt{3}}\\\\\\ h=\dfrac{100\cdot \big(1-\sqrt{3}\big)}{1^2-(\sqrt{3})^2}\\\\\\ h=\dfrac{100\cdot \big(1-\sqrt{3}\big)}{1-3}$

$h=\dfrac{100\cdot \big(1-\sqrt{3}\big)}{-2}\\\\\\ h=-50\cdot (1-\sqrt{3})\\\\\\ \therefore~~\boxed{\begin{array}{c}h=50\,(\sqrt{3}-1)\mathrm{~m} \end{array}}$


Bons estudos! :-)