Question

Determine o x na figura

Answer 1

Seja $AD=h$.

No triângulo $ABD$, temos que, $\text{tg}~60^{\circ}=\dfrac{AB}{AD}$.

Assim, $\sqrt{3}=\dfrac{x}{h}~~\texttt{(i)}$.

Além disso, no triângulo $ABC$, vemos que:

$\text{tg}~30^{\circ}=\dfrac{AB}{AC}$

Deste modo, $\dfrac{\sqrt{3}}{3}=\dfrac{x}{20+h}~~\texttt{(i)}$.

De $\texttt{(i)}$, tiramos que, $h=\dfrac{x}{\sqrt{3}}=\dfrac{x\sqrt{3}}{3}$.

De $\texttt{(ii)}$, temos $(20+h)\sqrt{3}=3x$, daí segue que, $h=\dfrac{3x\sqrt{3}}{3}-20$.

Logo:

$\dfrac{x\sqrt{3}}{3}=\dfrac{3x\sqrt{3}}{3}-20$

$x\sqrt{3}=3x\sqrt{3}-60$

$3x\sqrt{3}-x\sqrt{3}=60$

$2x\sqrt{3}=60$

$x\sqrt{3}=30$

Deste modo, $x=\dfrac{30}{\sqrt{3}}$ e obtemos $x=\dfrac{30\sqrt{3}}{3}$

Portanto, $x=10\sqrt{3}~\text{m}$.