In a right angled triangle ABC shown below, tan(angle ABC) = 1/2 and AB(hypotense)=square root of 20cm. Using simultaneous equations, find the length of AC(opposite) and BC(adjacent).
You are told that $\displaystyle \tan{\theta} = \frac{1}{2}$.
So $\displaystyle \frac{O}{A} = \frac{1}{2}$.
You also have, by Pythagoras' Theorem
$\displaystyle O^2 + A^2 = H^2$
$\displaystyle O^2 + A^2 = (\sqrt{20})^2$
$\displaystyle O^2 + A^2 = 20$.
So solve $\displaystyle \frac{O}{A} = \frac{1}{2}$ and $\displaystyle O^2 + A^2 = 20$ simultaneously.