Będziemy korzystać ze wzorów

$\cos \alpha =\frac{b^2+c^2-a^2}{2bc}$

$\cos \beta =\frac{a^2+c^2-b^2}{2ac}$

$\cos \gamma =\frac{b^2+a^2-c^2}{2ba}$

a) Oznaczmy

$a=5,b=14,c=15$  

Ze wzoru mamy

$\cos \alpha =\frac{{14}^2+{15}^2-5^2}{2\cdot 14\cdot 15}=\frac{196+225-25}{420}=\frac{396}{420}=\frac{33}{35}$  

$\cos \beta =\frac{5^2+{15}^2-{14}^2}{2\cdot 5\cdot 15}=\frac{25+225-196}{150}=\frac{54}{150}=\frac{9}{25}$  

$\cos \gamma =\frac{{14}^2+5^2-{15}^2}{2\cdot 5\cdot 14}=\frac{196+25-225}{140}=\frac{-4}{140}=-\frac{1}{35}$  

---

b) Oznaczmy

$a=9,b=10,c=7$  

Ze wzoru mamy

$\cos \alpha =\frac{{10}^2+7^2-9^2}{2\cdot 10\cdot 7}=\frac{100+49-81}{140}=\frac{68}{140}=\frac{17}{35}$  

$\cos \beta =\frac{9^2+7^2-{10}^2}{2\cdot 9\cdot 7}=\frac{81+49-100}{126}=\frac{30}{126}=\frac{5}{21}$  

$\cos \gamma =\frac{{10}^2+9^2-7^2}{2\cdot 10\cdot 9}=\frac{100+81-49}{180}=\frac{132}{180}=\frac{11}{15}$  

---

c) Oznaczmy

$a=7,b=13,c=16$   

Ze wzoru mamy

$\cos \alpha =\frac{{13}^2+{16}^2-7^2}{2\cdot 13\cdot 16}=\frac{169+256-49}{416}=\frac{376}{416}=\frac{47}{52}$  

$\cos \beta =\frac{7^2+{16}^2-{13}^2}{2\cdot 7\cdot 16}=\frac{49+256-169}{224}=\frac{136}{224}=\frac{17}{28}$  

$\cos \gamma =\frac{{13}^2+7^2-{16}^2}{2\cdot 13\cdot 7}=\frac{169+49-256}{182}=\frac{-38}{182}=-\frac{19}{91}$  

---

d) Oznaczmy

$a=1+\sqrt{2},b=2\sqrt{2},c=2$  

Ze wzoru mamy

$\cos \alpha =\frac{{\left(2\sqrt{2}\right)}^2+2^2-{\left(1+\sqrt{2}\right)}^2}{2\cdot 2\sqrt{2}\cdot 2}=\frac{8+4-\left(1+2\sqrt{2}+2\right)}{8\sqrt{2}}=\frac{12-3-2\sqrt{2}}{8\sqrt{2}}=\frac{9-2\sqrt{2}}{8\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{9\sqrt{2}-4}{16}$  

$\cos \beta =\frac{{\left(1+\sqrt{2}\right)}^2+2^2-{\left(2\sqrt{2}\right)}^2}{2\cdot \left(1+\sqrt{2}\right)\cdot 2}=\frac{\left(1+2\sqrt{2}+2\right)+4-8}{4\left(1+\sqrt{2}\right)}=\frac{3+2\sqrt{2}-4}{4\left(1+\sqrt{2}\right)}=\frac{2\sqrt{2}-1}{4\left(1+\sqrt{2}\right)}\cdot \frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{\left(2\sqrt{2}-1\right)\left(1-\sqrt{2}\right)}{4\left(1-2\right)}=\frac{2\sqrt{2}-4-1+\sqrt{2}}{-4}=-\frac{3\sqrt{2}-5}{4}=\frac{5-3\sqrt{2}}{4}$   

$\cos \gamma =\frac{{\left(2\sqrt{2}\right)}^2+{\left(1+\sqrt{2}\right)}^2-2^2}{2\cdot 2\sqrt{2}\cdot \left(1+\sqrt{2}\right)}=\frac{8+\left(1+2\sqrt{2}+2\right)-4}{4\left(\sqrt{2}+2\right)}=\frac{4+3+2\sqrt{2}}{4\left(\sqrt{2}+2\right)}=\frac{7+2\sqrt{2}}{4\left(\sqrt{2}+2\right)}\cdot \frac{\sqrt{2}-2}{\sqrt{2}-2}=\frac{\left(7+2\sqrt{2}\right)\left(\sqrt{2}-2\right)}{4\left(2-4\right)}=\frac{7\sqrt{2}-14+4-4\sqrt{2}}{-8}=-\frac{3\sqrt{2}-10}{8}=\frac{10-3\sqrt{2}}{8}$