Rysunek: 

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a)

O podanym trójkącie wiemy, że

• $\alpha ={60}^{\circ }$  
• $\beta ={45}^{\circ }$  
• $c=2\sqrt{3}$  

zatem

$\gamma ={180}^{\circ }-\left({60}^{\circ }+{45}^{\circ }\right)={180}^{\circ }-{105}^{\circ }={75}^{\circ }$  

Korzystając z twierdzenia sinusów mamy: 

$\frac{a}{\sin \alpha }=\frac{c}{\sin \gamma }$  

$\frac{a}{\sin {60}^{\circ }}=\frac{2\sqrt{3}}{\sin {75}^{\circ }}\mid \cdot \sin {60}^{\circ }$  

$a=\frac{2\sqrt{3}\cdot \sin {60}^{\circ }}{\sin {75}^{\circ }}$  

$a\approx \frac{2\sqrt{3}\cdot \frac{\sqrt{3}}{2}}{0,966}$  

$a\approx \frac{3}{0,966}$  

$\underline{\underline{a\approx 3,1}}$  

Korzystając z twierdzenia sinusów mamy: 

$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }$  

$\frac{3,1}{\sin {60}^{\circ }}\approx \frac{b}{\sin {45}^{\circ }}\mid \cdot \sin {45}^{\circ }$  

$b\approx \frac{3,1\cdot \sin {45}^{\circ }}{\sin {60}^{\circ }}$  

$b\approx \frac{3,1\cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}$  

$b\approx 3,1\cdot \frac{\sqrt{2}}{2}\cdot \frac{2}{\sqrt{3}}$  

$\underline{\underline{b\approx 2,5}}$  

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b)

O podanym trójkącie wiemy, że

• $\alpha ={45}^{\circ }$  
• $\beta ={15}^{\circ }$  
• $b=6$  

zatem

$\gamma ={180}^{\circ }-\left({45}^{\circ }+{15}^{\circ }\right)={180}^{\circ }-{60}^{\circ }={120}^{\circ }$  

Korzystając z twierdzenia sinusów mamy: 

$\frac{a}{\sin \alpha }=\frac{b}{\sin \beta }$  

$\frac{a}{\sin {45}^{\circ }}=\frac{6}{\sin {15}^{\circ }}\mid \cdot \sin {45}^{\circ }$  

$a=\frac{6\cdot \sin {45}^{\circ }}{\sin {15}^{\circ }}$  

$a\approx \frac{6\cdot \frac{\sqrt{2}}{2}}{0,259}$  

$\underline{\underline{a\approx 16,4}}$  

Korzystając z twierdzenia sinusów mamy: 

$\frac{a}{\sin \alpha }=\frac{c}{\sin \gamma }$  

$\frac{16,4}{\sin {45}^{\circ }}\approx \frac{c}{\sin {120}^{\circ }}\mid \cdot \sin {120}^{\circ }$  

$c\approx \frac{16,4\cdot \sin {120}^{\circ }}{\sin {45}^{\circ }}$  

$c\approx \frac{16,4\cdot \sin \left({90}^{\circ }+{30}^{\circ }\right)}{\sin {45}^{\circ }}$  

$c\approx \frac{16,4\cdot \cos {30}^{\circ }}{\sin {45}^{\circ }}$  

$c\approx \frac{16,4\cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}$  

$c\approx 16,4\cdot \frac{\sqrt{3}}{2}\cdot \frac{2}{\sqrt{2}}$  

$\underline{\underline{c\approx 20,1}}$