Question

Considerando a figura abaixo, determine a medida, aproximada de MN

Answer 1

$\text{MN}=\text{MQ+NQ }$

Façamos trigonometria nos respectivos triângulos retângulos :

1)

$\displaystyle \Delta_{\text{MQR}} \to \text{Tg}(60^{\circ}) = \frac{40 }{\text{MQ}}$

$\displaystyle \Delta_{\text{MQR}} \to \text{MQ} = \frac{40 }{\text{Tg}(60^{\circ})}$

$\displaystyle \Delta_{\text{MQR}} \to \text{MQ} = \frac{40 }{\sqrt3}$

$\displaystyle \Delta_{\text{MQR}} \to \text{MQ}\approx 23,09$

2)

$\displaystyle \Delta_{\text{NQP}} \to \text{Cos}(30^{\circ}) = \frac{\text{NQ}}{12}$

$\displaystyle \Delta_{\text{NQP}} \to \text{NQ} =12.\text{Cos}(30^{\circ})$

$\displaystyle \Delta_{\text{NQP}} \to \text{NQ} =12.\frac{\sqrt3}{2}$

$\displaystyle \Delta_{\text{NQP}} \to \text{NQ} \approx 10,39$

3)

$\text{MN}=\text{MQ+NQ }$

$\text{MN}\approx 23,09+10,39$

$\huge\boxed{\text{MN}\approx 33,5 \ \text{cm}} \checkmark$