Answer:
[tex]\sin x = \frac{1}{\sqrt{10} }[/tex] and [tex]\sin y = \frac{3}{\sqrt{10} }[/tex]
[tex]\cos x = \frac{3}{\sqrt{10} }[/tex] and [tex]\cos y = \frac{1}{\sqrt{10} }[/tex]
[tex]\tan x = \frac{1}{3}[/tex] and [tex]\tan y = 3[/tex]
Step-by-step explanation:
In a right triangle , let α be an angle of the triangle.
- [tex]\sin \alpha = \frac{side \: opposite \: to \: \alpha }{hypotenuse\: of \: the \: triangle}[/tex]
- [tex]\cos \alpha = \frac{side \: adjacent \: to \: \alpha }{hypotenuse \: of \: the \: triangle }[/tex]
- [tex]\tan \alpha = \frac{side \: opposite \: to \: \alpha }{side \: adjacent \: to \: \alpha }[/tex]
Using these properties , lets solve this question.
1) Let α (alpha) be x
- [tex]\sin x = \frac{1}{\sqrt{10} }[/tex]
- [tex]\cos x = \frac{3}{\sqrt{10} }[/tex]
- [tex]\tan x = \frac{1}{3}[/tex]
2) Let α (alpha) be y.
- [tex]\sin y = \frac{3}{\sqrt{10} }[/tex]
- [tex]\cos y = \frac{1}{\sqrt{10} }[/tex]
- [tex]\tan y = \frac{3}{1} = 3[/tex]
