Question

Um homem ve um predio sob um angulo de 30°, apos andar 100m, agora ele ve o mesmo predio sob um angulo de 45°. Qual a altura do predio?
Valendo 55

Answer 1

Explicação passo-a-passo:

Trigonômetria no triângulo rectângulo:

$\mathsf{\sin(30°)~=~\dfrac{h}{100+y} }$

$\mathsf{\dfrac{1}{2}~=~\dfrac{h}{100+y} }$

$\boxed{\mathsf{h~=~\dfrac{100+y}{2} }}}}$

$\mathsf{\sin(45°)~=~\dfrac{h}{y} }$

$\mathsf{\dfrac{\sqrt{2}}{2}~=~\dfrac{\frac{100+y}{2}}{y} }$

$\mathsf{\dfrac{y\sqrt{2}}{\cancel{2}}~=~\dfrac{100+y}{\cancel{2}} }$

$\mathsf{y\sqrt{2}~=~100+y }$

$\mathsf{y\sqrt{2}-y~=~100} }$

$\mathst{y\Big(\sqrt{2}-1\Big)~=~100 }$

$\mathsf{y~=~\dfrac{100}{\sqrt{2}-1} }$

Racionalizando o denominador ter-se-á:

$\mathsf{y~=~\dfrac{100}{\sqrt{2}-1}\times\dfrac{\sqrt{2}+1}{\sqrt{2}+1} }$

$\mathsf{y~=~\dfrac{100\sqrt{2}+100}{2-1} }$

$\boxed{\mathsf{y~=~100\sqrt{2}+100 }}}}$

Logo teremos que:

$\mathsf{h~=~\dfrac{100+y}{2} }$

$\mathsf{h~=~\dfrac{100+100\sqrt{2}+100}{2} }$

$\mathsf{h~=~\dfrac{200+100\sqrt{2}}{2} }$

$\mathsf{h~=~\dfrac{200}{2}+\dfrac{\cancel{100}\sqrt{2}}{\cancel{2}} }$

$\boxed{\boxed{\mathsf{altura(h)~=~100+50\sqrt{2}m }}}}$$\checkmark$

Espero ter ajudado bastante!)