Question

Помогите пожалуйста решить Соч по геометрии 2,3,4и5. ​

Answer 1

$2)\; \; \frac{1-sin^2a}{cos^2a}-sin^2a=\frac{cos^2a}{cos^2a}-sin^2a=1-sin^2a=cos^2a\\\\\\3)\; \; a=5\; ,\; \; b=8\; \; \to \; \; c=\sqrt{5^2+8^2}=\sqrt{89}\\\\a=5\; ,\; \; c=8\; \; \to \; \; b=\sqrt{8^2-5^2}=\sqrt{39}\\\\\\4)\; \; cosa=\frac{15}{17}\\\\sina=\pm \sqrt{1-cos^2a}=\pm \sqrt{1-\frac{225}{289}}=\pm \sqrt{\frac{64}{289}}=\pm \frac{8}{17}\\\\tga=\frac{sina}{cosa}=\pm \frac{8}{17}\cdot \frac{17}{15}=\pm \frac{8}{15}\\\\ctga=\frac{1}{tga}=\pm \frac{15}{8}$

$5)\; \; AC=4\sqrt3\; ,\; \; BD=4\; \; \Rightarrow \\\\AO=OC=2\sqrt3\; ,\; \; BO=OD=2\\\\tg\angle OAB=\frac{BO}{AO}=\frac{2}{2\sqrt3}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}\; \; \to \; \; \angle OAB=30^\circ \\\\\angle BAD=2\cdot \angle OAB=60^\circ \; \; ,\; \; \angle BCD=\angle BAD=60^\circ \\\\\angle ABC=\angle ADC=180^\circ -60^\circ =120^\circ$