Oznaczmy:

$\alpha$   - miara kąta między bokami  $a$   i  $b$  

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$\text{a)}$  

$P=\frac{1}{2}ab\sin \alpha$  

$12=\frac{1}{2}\cdot 4\cdot 10\cdot \sin \alpha$      

$12=20\sin \alpha$  

$\frac{12}{20}=\sin \alpha$  

$\frac{3}{5}=\sin \alpha$  

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\left(\frac{3}{5}\right)}^2+{\cos }^2\alpha =1$  

$\frac{9}{25}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{16}{25}$  

$\cos \alpha =\frac{4}{5}\text{lub}\cos \alpha =-\frac{4}{5}$  

 

Przypadek I.

$\cos \alpha =\frac{4}{5}$  

Wtedy mamy:

$c^2=a^2+b^2-2ab\cdot \cos \alpha$  

$c^2=4^2+{10}^2-2\cdot 4\cdot 10\cdot \frac{4}{5}$  

$c^2=16+100-64$  

$c^2=52$  

$c=\sqrt{52}$  

$c=2\sqrt{13}$  

 

Przypadek II.

$\cos \alpha =-\frac{4}{5}$  

Wtedy mamy:

$c^2=a^2+b^2-2ab\cdot \cos \alpha$  

$c^2=4^2+{10}^2-2\cdot 4\cdot 10\cdot \left(-\frac{4}{5}\right)$  

$c^2=16+100+64$  

$c^2=180$  

$c=\sqrt{180}$  

$c=6\sqrt{5}$  

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$\text{b)}$  

$P=\frac{1}{2}ab\sin \alpha$  

$10\sqrt{2}=\frac{1}{2}\cdot 5\cdot 6\cdot \sin \alpha$  

$10\sqrt{2}=15\sin \alpha$  

$\frac{10\sqrt{2}}{15}=\sin \alpha$  

$\frac{2\sqrt{2}}{3}=\sin \alpha$  

Korzystając z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\left(\frac{2\sqrt{2}}{3}\right)}^2+{\cos }^2\alpha =1$  

$\frac{8}{9}+{\cos }^2\alpha =1$  

${\cos }^2\alpha =\frac{1}{9}$  

$\cos \alpha =\frac{1}{3}\text{lub}\cos \alpha =-\frac{1}{3}$  

 

Przypadek I.

$\cos \alpha =\frac{1}{3}$  

Wtedy mamy:

$c^2=a^2+b^2-2ab\cdot \cos \alpha$  

$c^2=5^2+6^2-2\cdot 5\cdot 6\cdot \frac{1}{3}$  

$c^2=25+36-20$  

$c^2=41$  

$c=\sqrt{41}$  

 

Przypadek II.

$\cos \alpha =-\frac{1}{3}$  

Wtedy mamy:

$c^2=a^2+b^2-2ab\cdot \cos \alpha$  

$c^2=5^2+6^2-2\cdot 5\cdot 6\cdot \left(-\frac{1}{3}\right)$  

$c^2=25+36+20$  

$c^2=81$  

$c=9$