Trigonometry find area of triangle Triangle ABC is right angeled with right angle at corner C and angle a at corner A. Calculate triangle area if we know that c=|AB|=10 and tan a = 3/2
I get side |AC| = 2
side |CB| = 3
and side |AB| = 10 but this can not be case because it's not pythagoras definition then 
Greetings from RUssia
Vladimir Vlostok

$A = 0.5 \cdot |AB||AC| \sin(a)$
$\tan(a)=\frac{\sin(a)}{\cos(a)}=\frac{\sin(a)}{\sqrt{1-\sin^2(a)}}$. Solve this equaiton for $\sin(a)$:
$\sin(a)=\frac{\tan(a)}{\sqrt{1+\tan^2(a)}}=\frac{3/2}{\sqrt{1+9/4}}=\frac{3}{\sqrt{13}}$
 and plug it into the previous equation. Note, that the previous statements are only true for $0\leq a \leq \pi/2$, which is fulfilled.
