Даны точки A(134;-887). B(133:-892), C(143:-890) найдите тангенс угла АВС

Решите задачу:

$BA=(134-133;-887+892)=(1;5)$
$|AB|= \sqrt{1^2+5^2} = \sqrt{26}$
$BC=(143-133;-890+892)=(10;2)$
$|BC|= \sqrt{10^2+2^2}= \sqrt{104}=2 \sqrt{26}$
$cos(ABC)= \frac{BA*BC}{|BA|*|BC|} = \frac{1*10+5*2}{ \sqrt{26}*2 \sqrt{26} } = \frac{20}{2*26} = \frac{5}{13}$
$sin(ABC)= \sqrt{1-cos^2(ABC)} = \sqrt{1- \frac{25}{169} } = \sqrt{ \frac{144}{169} } = \frac{12}{13}$
$tg(ABC)= \frac{sin(ABC)}{cos(ABC)} = \frac{12}{13} : \frac{5}{13} = \frac{12}{5} =2,4$