## Elementary Geometry for College Students (6th Edition)

cos α = $\frac{2}{sqrt of 13}$ , cos β = $\frac{3}{sqrt of 13}$
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 = $\frac{3}{sqrt of 13}$