Question

in triunghiul ABC AB=5 AC=12 BC=13 calculati cosB

Answer 1

ΔABC dreptunghic
AB = 5
AC = 12
BC = 13
cos(B) = ?
$cos(B)= \frac{AB}{BC}= \frac{5}{13}$

Answer 2

$\displaystyle AB=5;~AC=12;~BC=13;~cos~B=? \\ \boxed{cos~B= \frac{a^2+c^2-b^2}{2ac} } \\ AB=c=5;~AC=b=12;~BC=a=13 \\ cos~B= \frac{13^2+5^2-12^2}{2 \cdot 13 \cdot 5} = \frac{169+25-144}{130} = \frac{50}{130} ^{(10} = \frac{5}{13} \\ cos~B= \frac{5}{13}$