Suma miar kątów trójkąta wynosi  ${180}^{\circ }$ .

$\text{I}.$  

${180}^{\circ }-\left({90}^{\circ }+{60}^{\circ }\right)={180}^{\circ }-{150}^{\circ }={30}^{\circ }$  

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$\text{II}.$  

${180}^{\circ }-\left({90}^{\circ }+{45}^{\circ }\right)={180}^{\circ }-{135}^{\circ }={45}^{\circ }$  

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$\text{III}.$  

${180}^{\circ }-\left({90}^{\circ }+{70}^{\circ }\right)={180}^{\circ }-{160}^{\circ }={20}^{\circ }$  

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$\text{IV}.$  

${180}^{\circ }-\left({120}^{\circ }+{20}^{\circ }\right)={180}^{\circ }-{140}^{\circ }={40}^{\circ }$  

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$\text{V}.$  

${180}^{\circ }-\left({55}^{\circ }+{55}^{\circ }\right)={180}^{\circ }-{110}^{\circ }={70}^{\circ }$  

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$\text{VI}.$  

${180}^{\circ }-\left({80}^{\circ }+{40}^{\circ }\right)={180}^{\circ }-{120}^{\circ }={60}^{\circ }$