Если sinA=4/5,и cosB=12/13 и A и В углы острые,то найдите 13sin(A+B).

$\sin \alpha=\frac{4}{5}\\ \\ \cos^2 \alpha+\sin^2 \alpha=1\\\\ \cos^2 \alpha= 1-\sin^2 \alpha\\ \\ \cos \alpha=\sqrt{1-\sin^2 \alpha}=\sqrt{1-(\frac{4}{5})^2}=\sqrt{1-\frac{16}{25}} =\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5} \\\\ \cos \beta=\frac{12}{13}\\ \\ \cos^2\beta+\sin^2 \beta=1\\\\ \cos^2\beta= 1-\sin^2 \beta\\ \\ \sin \beta=\sqrt{1-\cos^2 \beta}=\sqrt{1-(\frac{12}{13})^2}=\sqrt{1-\frac{144}{169}} =\sqrt{\frac{169-144}{169}}=\\\\=\sqrt{\frac{25}{169}}=\frac{5}{13}$

$13\sin(\alpha +\beta )=13(\sin\alpha \cdot\cos\beta +\cos\alpha \cdot\sin\beta )=13(\frac{4}{5} \cdot\frac{12}{13}+\frac{3}{5} \cdot\frac{5}{13})=\\ \\ =\frac{48}{5}+3=9,6+3=12,6$

Ответ: $12,6$