Question

помогите решить задачи 18 И 19​

Answer 1

$18)\\sin(\pi +\alpha )=-sin\alpha \\\\sin \alpha =\frac{AC}{AB} \\\\-sin \alpha =-\frac{AC}{AB} \\\\AB^{2} =BC^2+AC^2\\\\AB^2=64+36=100\\\\AB=10\\\\-sin\alpha =-\frac{6}{10} =-0.6\\\\Otvet: B)\\\\19)\\sin(\frac{\pi }{2} -\alpha )+sin(\frac{3\pi }{2} -\beta )= cos\alpha - cos\beta\\\\cos\alpha =\frac{AB}{AC} \\\\cos\beta =\frac{AD}{AC}\\ \\AC^2=AB^2+BC^2\\\\AC^2=64+36=100\\\\AC=10\\\\cos\alpha =\frac{AB}{AC} =\frac{8}{10} =0.8\\\\cos\beta =\frac{AD}{AC}=\frac{6}{10} =0.6$

$sin(\frac{\pi }{2} -\alpha )+sin(\frac{3\pi }{2} -\beta )= cos\alpha - cos\beta=0.8-0.6=0.2$