Question

Um triângulo retângulo tem um angulo agudo α tal que tg α = √5/2. Calcule os valores de sen α e cos β.

Answer 1

Segundo as relações trigonométricas:

sec²x = 1 + tg²x
sec²x = 1 + (√5/2)²
sec²x = 1 + 5/4 = 9/4
secx = 3/2

sec x = 1/cosx
cos x = 2/3


Como cos²x+sen²x = 1
(2/3)² + sen² x = 1
sen²x = 1 - 4/9
sen²x = 5/9

senx = √5/3

Answer 2

Pela relação fundamental da trigonometria:
$\sin^2\alpha+\cos^2\alpha=1\\\\\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha}=\dfrac{1}{\cos^2\alpha}\\\\\tan^2\alpha+1=\dfrac{1}{\cos^2\alpha}$

Através desta última relação podemos calcular o cosseno de α sabendo a tangente de α.
$\tan^2\alpha+1=\dfrac{1}{\cos^2\alpha}\\\\\left(\dfrac{\sqrt{5}}{2}\right)^2+1=\dfrac{1}{\cos^2\alpha}\\\\\dfrac{5}{4}+1=\dfrac{1}{cos^2\alpha}\\\\\cos^2\alpha=\dfrac{1}{\dfrac{5}{4}+1}\\\\\cos^2\alpha=\dfrac{1}{\dfrac{5+4}{4}}\\\\\cos^2\alpha=\dfrac{4}{9}\\\\\boxed{\boxed{\cos\alpha=\dfrac{2}{3}}}$

Pela própria relação fundamental da trigonometria, calcularemos o seno de α.
$\sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+\left(\dfrac{2}{3}\right)^2=1\\\\\sin^2\alpha+\dfrac{4}{9}=1\\\\\sin^2\alpha=1-\dfrac{4}{9}\\\\\sin^2\alpha=\dfrac{9-4}{9}\\\\\boxed{\boxed{\sin\alpha=\dfrac{\sqrt5}{3}}}$