*Obs: Usei derivada, foi o jeito mais rápido que pensei para resolver i) $\frac{a}{b} = 3 \therefore a = 3b$ ii) $[ABC] = \frac{1}{2}a.bsen(\theta) = \frac{1}{2}bh \therefore h = \frac{3b^2sen(\theta)}{32}$ iii) $Lei\ dos\ cossenos:$ $32^2 = 9b^2 + b^2 - 6b^2.cos(\theta) \therefore b^2 = \frac{32^2}{10 - 6cos(\theta)}$ $h = \frac{3sen(\theta)\frac{32^2}{10 - 6cos(\theta))}}{32} \therefore h(\theta) = \frac{96sen(\theta)}{10 - 6cos(\theta)}$ iv) $h'(\theta) = \frac{96cos(\theta)(10 -6cos(\theta)) - 96sen(\theta)(6 sen(\theta))}{(10 - 6cos(\theta))^2} = 0 $ $\therefore 960cos(\theta) - 96.6cos^2(\theta) - 96.6sen^2(\theta) = 0$ $\therefore 960cos(\theta) - 96.6 = 0 \Rightarrow cos(\theta) = \frac{3}{5} \therefore sen(\alpha) = \frac{4}{5}$ $\Leftrightarrow h = \frac{96.\frac{4}{5}}{10 - 6\frac{3}{5}} \therefore h = 12$