Răspuns:
Explicație pas cu pas:
[tex]sinB=\frac{AC}{BC} \Leftrightarrow \frac{3}{4}=\frac{AC}{10} \Rightarrow AC=\frac{3 \times 10}{4}=\frac{30}{4}=\frac{15}{2}cm \\ Aplicam \ Teorema \ lui \ Pitagora \: pentru \: a \: afla \: AB: \\ \ AB {}^{2} =BC{}^{2} -AC {}^{2} =10 {}^{2} -\Big(\frac{15}{2} \Big)^{2}=\frac{{}^{4)}100}{1}-\frac{225}{4} \\ \ AB^{2}=\frac{400}{4}-\frac{225}{4}=\frac{175}{4} \Rightarrow AB=\sqrt{\frac{175}{4}}=\frac{5\sqrt{7}}{2}cm \\ \ P_{\triangle ABC} =AB+BC+AC=\frac{5\sqrt{7}}{2}+\frac{{}^{2)}10}{1}+\frac{15}{2}=\frac{5\sqrt{7}+20+15}{2}=\red{\boxed{\frac{5\sqrt{7}+35}{2}cm}}[/tex]
