Do alto de um farol cuja altura é de 20 m avista se um navio sob ângulo de depressão de 30°. A que distância , aproximadamente,
o navio se acha do farol ?
1 answer:
Esta é a trigonometria . Se você desenhar uma linha a partir do topo da casa de luz para o barco, você terá a hypotonuse de um triângulo. Um truque é lembrar que este é um triângulo especial. É um triângulo 30-60-90 , que tem propriedades especiais mostradas na fixação abaixo . por isso sabemos que o lado adjacente que não é o hyposonuse é x√3 . Agora sabemos que x<span>√3 = 20
Solve for x.
x</span><span>√3=20
divide both sides by </span><span>√3.
x=20/(</span><span><span>√3)
</span>Try not to have square roots (</span><span><span>√)</span> in denomenator so multiply top and bottom by </span><span>√3 and get
x=(20</span><span>√3)/3
x is what we are looking for so the answer is </span>
20<span>√3 m </span><span>ou cerca de 34.64 m</span>
Solve for x.
x</span><span>√3=20
divide both sides by </span><span>√3.
x=20/(</span><span><span>√3)
</span>Try not to have square roots (</span><span><span>√)</span> in denomenator so multiply top and bottom by </span><span>√3 and get
x=(20</span><span>√3)/3
x is what we are looking for so the answer is </span>
20<span>√3 m </span><span>ou cerca de 34.64 m</span>
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