Answer:
[tex]q = 16 \sqrt{2} [/tex] and [tex]r = 16[/tex]
Step-by-step explanation:
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 \: traingle} [/tex]
- [tex] \tan( \alpha ) = \frac{side \: opposite \: to \: \alpha }{side \: adjacent \: to \: \alpha } [/tex]
where [tex] \alpha [/tex] is an angle of a triangle.
In the triangle given above , [tex] \alpha = 45° [/tex]. So,
1) [tex] \sin( 45) = \frac{16}{q} [/tex]
[tex] = > \frac{1}{ \sqrt{2} } = \frac{16}{q} [/tex]
[tex] = > q = 16 \sqrt{2} [/tex]
2) [tex] \tan(45) = \frac{16}{r} [/tex]
[tex] = > 1 = \frac{16}{r} [/tex]
[tex] = > r = 16[/tex]
