Answer
cos α = $\frac{2}{sqrt of 13}$ , cos β = $\frac{3}{sqrt of 13}$
Work Step by Step
Step 1: By Pythagoras theorem
$b^{2}$ +$3^{2}$ = square of ($\sqrt 13$)
$b^{2}$ + 9 = 13
$b^{2}$ = 13 -9 = 4
b = $\sqrt 4$ = 2
Step 2:
cos α =$\frac{ lengthofadjacent}{lengthofhypotenuse}$
cos α = $\frac{2}{sqrt of 13}$
Similarly cos β = $\frac{ lengthofadjacent}{lengthofhypotenuse}$
cos β = $\frac{3}{sqrt of 13}$
Therefore cos α = $\frac{2}{sqrt of 13}$ , cos β = $\frac{3}{sqrt of 13}$
