Question

Does anyone know the answer to this question

Answer 1

ANSWER

$\cos B = \frac{ \sqrt{3} }{3}$

EXPLANATION

The given triangle is a right triangle.

It was given that,

$a = 1$

and

$b = \sqrt{2}$

Using the Pythagoras Theorem, we can determine the value of c.

${c}^{2} = {( \sqrt{2} )}^{2} + {1}^{2}$

${c}^{2} = 2 + 1$

${c}^{2} = 3$

$c = \sqrt{3}$

The ratio is the adjacent over the hypotenuse.

$\cos B = \frac{1 }{ \sqrt{3} }$

We rationalize to get:

$\cos B = \frac{ \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } = \frac{ \sqrt{3} }{3}$