ANSWER
[tex]\cos B = \frac{ \sqrt{3} }{3} [/tex]
EXPLANATION
The given triangle is a right triangle.
It was given that,
[tex]a = 1[/tex]
and
[tex]b = \sqrt{2} [/tex]
Using the Pythagoras Theorem, we can determine the value of c.
[tex] {c}^{2} = {( \sqrt{2} )}^{2} + {1}^{2} [/tex]
[tex] {c}^{2} = 2 + 1[/tex]
[tex]{c}^{2} = 3[/tex]
[tex]c = \sqrt{3} [/tex]
The ratio is the adjacent over the hypotenuse.
[tex]\cos B = \frac{1 }{ \sqrt{3} } [/tex]
We rationalize to get:
[tex]\cos B = \frac{ \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{ \sqrt{3} }{3} [/tex]
