$\text{a)}$  

Korzystając z sumy miar kątów w trójkącie:

$\alpha ={180}^{\circ }-\beta =\gamma ={180}^{\circ }-{30}^{\circ }-{45}^{\circ }={105}^{\circ }$  

 

Korzystając z tw. sinusów otrzymujemy: 

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

$\frac{a}{\sin {105}^{\circ }}=\frac{\sqrt{2}}{\sin {45}^{\circ }}$  

$\frac{a}{\sin \left({90}^{\circ }+{15}^{\circ }\right)}=\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}$  

$\frac{a}{\cos {15}^{\circ }}=\sqrt{2}\cdot \frac{2}{\sqrt{2}}$  

$\frac{a}{\cos {15}^{\circ }}=2$  

Z ćwiczenia 1 ze strony 336 wiemy, że  $\cos {15}^{\circ }=\frac{\sqrt{6}+\sqrt{2}}{4}$  

$\frac{a}{\frac{\sqrt{6}+\sqrt{2}}{4}}=2$  

$a=2\cdot \frac{\sqrt{6}+\sqrt{2}}{4}$  

$a=\frac{\sqrt{6}+\sqrt{2}}{2}$  

 

$\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }$  

$\frac{b}{\sin {30}^{\circ }}=\frac{\sqrt{2}}{\sin {45}^{\circ }}$  

$\frac{b}{\frac{1}{2}}=\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}$  

$2b=2$  

$b=1$  

---

$\text{b)}$  

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

$\frac{5}{\sin {30}^{\circ }}=\frac{5\sqrt{3}}{\sin \beta }$  

$\frac{5}{\frac{1}{2}}=\frac{5\sqrt{3}}{\sin \beta }$  

$10=\frac{5\sqrt{3}}{\sin \beta }$  

$\sin \beta =\frac{5\sqrt{3}}{10}$  

$\sin \beta =\frac{\sqrt{3}}{2}$  

$\sin \beta =\sin {60}^{\circ }\vee \sin \beta ={120}^{\circ }$  

$\beta ={60}^{\circ }\vee \beta ={120}^{\circ }$  

 

Przypadek I.

$\beta ={60}^{\circ }$  

Wtedy mamy:

$\gamma ={180}^{\circ }-\alpha -\beta ={180}^{\circ }-{30}^{\circ }-{60}^{\circ }={90}^{\circ }$  

Zatem:

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

$\frac{5}{\frac{1}{2}}=\frac{c}{\sin {90}^{\circ }}$  

$10=\frac{c}{1}$  

$10=c$  

 

Przypadek II.

$\beta ={120}^{\circ }$  

Wtedy mamy:

$\gamma ={180}^{\circ }-\alpha -\beta ={180}^{\circ }-{30}^{\circ }-{120}^{\circ }={30}^{\circ }$  

Zatem:

 

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

$\frac{5}{\frac{1}{2}}=\frac{c}{\sin {30}^{\circ }}$  

$10=\frac{c}{\frac{1}{2}}$  

$5=c$  

---

$\text{c)}$  

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

$\frac{7}{\sin \alpha }=\frac{14}{\sin {60}^{\circ }}$  

$\frac{7}{\sin \alpha }=\frac{14}{\frac{\sqrt{3}}{2}}$  

$7\cdot \frac{\sqrt{3}}{2}=14\cdot \sin \alpha \mid :14$  

$\frac{\sqrt{3}}{4}=\sin \alpha$  

$0,433\approx \sin \alpha$  

${26}^{\circ }\approx \alpha$  

 

Korzystając z sumy miar kątów w trójkącie:

$\beta ={180}^{\circ }-\alpha -\beta \approx {180}^{\circ }-{26}^{\circ }-{60}^{\circ }={94}^{\circ }$  

Zatem mamy:

$\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }$  

$\frac{b}{\sin {94}^{\circ }}=\frac{14}{\sin {60}^{\circ }}$  

$\frac{b}{\sin \left({90}^{\circ }+4^{\circ }\right)}=\frac{14}{\frac{\sqrt{3}}{2}}$  

$\frac{b}{\cos 4^{\circ }}=\frac{28}{\sqrt{3}}$  

$\frac{b}{0,9976}\approx \frac{28}{\sqrt{3}}$  

$b\approx \frac{28}{1,73}\cdot 0,9976$  

$b\approx 16,146$  

---

$\text{d)}$  

Korzystając z sumy miar kątów w trójkącie:

$\gamma ={180}^{\circ }-\alpha -\beta ={180}^{\circ }-{150}^{\circ }-{10}^{\circ }={20}^{\circ }$  

 

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

$\frac{a}{\sin {150}^{\circ }}=\frac{8,7}{\sin {10}^{\circ }}$  

$\frac{a}{\sin \left({90}^{\circ }+{60}^{\circ }\right)}\approx \frac{8,7}{0,1736}$  

$\frac{a}{\cos {60}^{\circ }}\approx \frac{87}{1,736}$  

$\frac{a}{0,5}\approx \frac{87}{1,736}$  

$a\approx \frac{87}{1,736}\cdot 0,5$  

$a\approx 25,058$  

 

$\frac{b}{\sin \beta }=\frac{c}{\sin \gamma }$  

$\frac{8,7}{\sin {10}^{\circ }}=\frac{c}{\sin {20}^{\circ }}$  

$\frac{8,7}{0,1736}\approx \frac{c}{0,3420}$  

$\frac{8,7}{0,1736}\cdot 0,342\approx c$   

$17,139\approx c$