a)

Z twierdzenia Pitagorasa w trójkącie ABC mamy

$$\begin{gathered} {\left|BC\right|}^2+{\left|AC\right|}^2={\left|AB\right|}^2 \\ {\left|BC\right|}^2+15^2=25^2 \\ {\left|BC\right|}^2+225=625 \\ {\left|BC\right|}^2=400,\left|BC\right|>0 \\ \left|BC\right|=20 \end{gathered}$$

Stąd obliczmy

$$\begin{gathered} \mathrm{tg}\alpha =\frac{\left|BC\right|}{\left|AC\right|}=\frac{20}{15}=\frac{4}{3} \\ \mathrm{tg}\beta =\frac{\left|AC\right|}{\left|BC\right|}=\frac{15}{20}=\frac{3}{4} \end{gathered}$$

---

b)

Z twierdzenia Pitagorasa w trójkącie ABC mamy

$$\begin{gathered} {\left|AB\right|}^2+{\left|BC\right|}^2={\left|AC\right|}^2 \\ {\left|AB\right|}^2+5^2=13^2 \\ {\left|AB\right|}^2+25=169 \\ {\left|AB\right|}^2=144,\left|AB\right|>0 \\ \left|AB\right|=12 \end{gathered}$$

Stąd obliczmy

$$\begin{gathered} \mathrm{tg}\alpha =\frac{\left|BC\right|}{\left|AB\right|}=\frac{5}{12} \\ \mathrm{tg}\beta =\frac{\left|AB\right|}{\left|BC\right|}=\frac{12}{5} \end{gathered}$$

---

c)

Z twierdzenia Pitagorasa w trójkącie ABC mamy

$$\begin{gathered} {\left|BC\right|}^2+{\left|AC\right|}^2={\left|AB\right|}^2 \\ {\left|BC\right|}^2+(4\sqrt{3}{)}^2=8^2 \\ {\left|BC\right|}^2+48=64 \\ {\left|BC\right|}^2=16,\left|BC\right|>0 \\ \left|BC\right|=4 \end{gathered}$$

Stąd obliczmy

$$\begin{gathered} \mathrm{tg}\alpha =\frac{\left|BC\right|}{\left|AC\right|}=\frac{4}{4\sqrt{3}}=\frac{\sqrt{3}}{3} \\ \mathrm{tg}\beta =\frac{\left|AC\right|}{\left|BC\right|}=\frac{4\sqrt{3}}{4}=\sqrt{3} \end{gathered}$$

---

d)

Z twierdzenia Pitagorasa w trójkącie ABC mamy

$$\begin{gathered} {\left|AB\right|}^2+{\left|BC\right|}^2={\left|AC\right|}^2 \\ {\left|AB\right|}^2+(\sqrt{2}-1{)}^2=(\sqrt{6}{)}^2 \\ {\left|AB\right|}^2+2-2\sqrt{2}+1=6 \\ {\left|AB\right|}^2=3+2\sqrt{2},\left|AB\right|>0 \\ \left|AB\right|=\sqrt{3+2\sqrt{2}} \end{gathered}$$

Przekształćmy

$\left|AB\right|=\sqrt{3+2\sqrt{2}}=\sqrt{1+2\sqrt{2}+2}=\sqrt{{\left(1+\sqrt{2}\right)}^2}=\left|1+\sqrt{2}\right|=1+\sqrt{2}$

Stąd obliczmy

$$\begin{gathered} \mathrm{tg}\alpha =\frac{\left|BC\right|}{\left|AB\right|}=\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1{)}^2}{2-1}=2-2\sqrt{2}+1=3-2\sqrt{2} \\ \mathrm{tg}\beta =\frac{\left|AB\right|}{\left|BC\right|}=\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1{)}^2}{2-1}=2+2\sqrt{2}+1=3+2\sqrt{2} \end{gathered}$$