Answer:
1) [tex] \cos(A) = 0.948(appx)[/tex]
2) [tex] \sin(A) = 0.316(appx)[/tex]
3) [tex] \tan(A) = 0.333(appx)[/tex]
Step-by-step explanation:
Using Pythagorean theorem ,
Hypotenuse =
[tex] \sqrt{ {6}^{2} + {2}^{2} } = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} [/tex]
We know that
- [tex] \sin( \alpha ) = \frac{side \: opposite \: to \: \alpha }{hypotenuse \: of \: triangle} [/tex]
- [tex] \cos( \alpha ) = \frac{side \: adjacent \: to \: \alpha }{hypotenuse \: of \: triangle} [/tex]
- [tex] \tan( \alpha ) = \frac{side \: opposite \: to \: \alpha }{side \: adjacent \: to \: \alpha } [/tex]
Where [tex] \ alpha [/tex] is an angle of tge triangle.
In the triangle given above , [tex] \ alpha = angle \: A [/tex] . So,
1) [tex] \cos(A) = \frac{6}{2 \sqrt{10} } = \frac{3}{ \sqrt{10} } = 0.948(appx)[/tex]
2) [tex] \sin(A) = \frac{2}{2 \sqrt{10} } = \frac{1}{ \sqrt{10} } = 0.316(appx)[/tex]
3) [tex] \tan(A) = \frac{2}{6} = \frac{1}{3} = 0.333(appx)[/tex]
