Question

найдите пожалуйста стороны ​

Answer 1

Ответ:

$\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\\\\\\\angle {C}=180^\circ -(60^\circ +75^\circ )=45^\circ \\\\\\\dfrac{4}{sin60^\circ }=\dfrac{c}{sin45^\circ }\ \ \to \ \ \ c=\dfrac{4\cdot sin45^\circ }{sin60^\circ }=\dfrac{4\cdot \sqrt2/2}{\sqrt3/2}=\dfrac{4\sqrt2}{\sqrt3}=\dfrac{4\sqrt6}{3}\\\\\\\dfrac{4}{sin60^\circ }=\dfrac{a}{sin75^\circ }\ \ \to \ \ \ a=\dfrac{4\cdot sin75^\circ }{sin60^\circ }=\dfrac{2\cdot (\sqrt6+\sqrt2)}{\sqrt3}=\dfrac{2\sqrt2\cdot (\sqrt3+1)}{3}$

$\star \ \ sin75^\circ =sin(45^\circ +30^\circ )=sin45^\circ \cdot cos30^\circ +cos45^\circ \cdot sin30^\circ =\\\\=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt6+\sqrt2}{4} \ \ \star$