Question

срочно нужно найти треугольник дам 85б​

Answer 1

$\displaystyle\bf\\<A+<B+<C=180^\circ\\\\<C=180^\circ-(60^\circ+75^\circ)=180^\circ-135^\circ=45^\circ\\\\\\\frac{AC}{Sin<B} =\frac{AB}{Sin<C} \\\\\\\frac{4}{Sin60^\circ} =\frac{AB}{Sin45^\circ } \\\\\\AB=\frac{4\cdot Sin45^\circ}{Sin60^\circ } =\frac{4\cdot\frac{\sqrt{2} }{2} }{\frac{\sqrt{3} }{2} } =\frac{4\sqrt{2} }{\sqrt{3} }=\frac{4\sqrt{6} }{3} \\\\\\\frac{BC}{Sin<A} =\frac{AC}{Sin<B} \\\\\\\frac{BC}{Sin75^\circ} =\frac{4}{Sin60^\circ }$

$\displaystyle\bf\\BC=\frac{4\cdot Sin75^\circ}{Sin60^\circ } =\frac{4\cdot\frac{\sqrt{6} +\sqrt{2} }{4} }{\frac{\sqrt{3} }{2} } =\frac{2(\sqrt{6} +\sqrt{2} )}{\sqrt{3} } \\\\\\Sin75^\circ=Sin(30^\circ+45^\circ)= Sin30^\circ\cdot Cos45^\circ+Sin45^\circ Cos30^\circ=\\\\\\=\frac{1}{2} \cdot\frac{\sqrt{2} }{2} +\frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2} =\frac{\sqrt{2} }{4}+\frac{\sqrt{6} }{4} =\frac{\sqrt{2}+\sqrt{6} }{4}$