Question

Por favor preciso de ajuda.
Mas com resolução.

Answer 1

$tg \alpha = \dfrac{\text{cateto oposto}}{\text{cateto adjacente}}$

$tg 30^{o} = \dfrac{\text{H}}{\text{X}}$ ⇒ $\text{X} = \dfrac{\text{H}}{tg 30^{o}}$ ⇒ $\text{X} = \dfrac{\text{H}}{ \frac{ \sqrt{3} }{3} } = \dfrac{3\text{H}}{ \sqrt{3} } = \dfrac{3\sqrt{3}\text{ H}}{ 3 } =\sqrt{3}\text{ H}$

$tg 60^{o} = \dfrac{\text{H}}{\text{X-30 }}$ ⇒ $\text{X-30 } = \dfrac{\text{H}}{tg 60^{o}}= \dfrac{\text{H}}{ \sqrt{3} }= \dfrac{ \sqrt{3} \text{H}}{ 3}$

$\text{X } = \dfrac{ \sqrt{3} \text{H}}{ 3}+30= \dfrac{ \sqrt{3} \text{H}+90}{ 3}$

$\sqrt{3}\text{ H}=\dfrac{ \sqrt{3} \text{H}+90}{ 3}$

$3\sqrt{3} \text{H}=\sqrt{3} \text{H}+90$

$3\sqrt{3} \text{H}-\sqrt{3} \text{H}=90$

$2\sqrt{3} \text{H}=90$

$\text{H}= \dfrac{90}{2\sqrt{3} }= \dfrac{45}{\sqrt{3} }= \dfrac{45 \sqrt{3} }{3}=15 \sqrt{3}m$