Question

Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)

Answer 1

Answer:

$\sin x = \frac{1}{\sqrt{10} }$  and $\sin y = \frac{3}{\sqrt{10} }$

$\cos x = \frac{3}{\sqrt{10} }$  and $\cos y = \frac{1}{\sqrt{10} }$

$\tan x = \frac{1}{3}$  and $\tan y = 3$

Step-by-step explanation:

In a right triangle , let α be an angle of the triangle.

• $\sin \alpha = \frac{side \: opposite \: to \: \alpha }{hypotenuse\: of \: the \: triangle}$

• $\cos \alpha = \frac{side \: adjacent \: to \: \alpha }{hypotenuse \: of \: the \: triangle }$

• $\tan \alpha = \frac{side \: opposite \: to \: \alpha }{side \: adjacent \: to \: \alpha }$

Using these properties , lets solve this question.

1) Let α (alpha) be x

• $\sin x = \frac{1}{\sqrt{10} }$

• $\cos x = \frac{3}{\sqrt{10} }$

• $\tan x = \frac{1}{3}$

2) Let α (alpha) be y.

• $\sin y = \frac{3}{\sqrt{10} }$

• $\cos y = \frac{1}{\sqrt{10} }$

• $\tan y = \frac{3}{1} = 3$