Пусть [tex]s_n=\sin n^{\circ}[/tex] и [tex]$c_n=\cos n^{\circ}$[/tex], тогда

[tex]s_{80}s_{65}s_{35}=s_{80}\cfrac{c_{30}-c_{100}}2=\cfrac{(s_{110}+s_{50})-(s_{180}-s_{20})}4[/tex]

А значит

[tex]\cfrac{96s_{80}s_{65}s_{35}}{s_{110}+s_{50}+s_{20}}=\cfrac{96}4=24[/tex]

Ответ:

Применяем формулы произведение синусов и произведение синуса на косинус :

  [tex]\boldsymbol{sin\alpha \cdot sin\beta =\dfrac{1}{2}\cdot \Big(cos(\alpha -\beta )-cos(\alpha +\beta )\Big)}\ \ ,\\\\\boldsymbol{sin\alpha \cdot cos\beta =\dfrac{1}{2}\cdot \Big(sin(\alpha +\beta )+sin(\alpha -\beta )\Big)\ \ .}[/tex]                  

[tex]\bf \dfrac{96\cdot sin80^\circ \cdot sin65^\circ \cdot sin35^\circ }{sin20^\circ +sin50^\circ +sin110^\circ }=\dfrac{96\cdot sin80^\circ \cdot (sin65^\circ \cdot sin35^\circ )}{sin20^\circ +sin50^\circ +sin110^\circ }=\\\\\\=\dfrac{96\cdot sin80^\circ \cdot \dfrac{1}{2}\, (cos30^\circ -cos100^\circ )}{sin20^\circ +sin50^\circ +sin110^\circ }=\dfrac{48\cdot (sin80^\circ \cdot cos30^\circ -sin80^\circ \cdot cos100^\circ )}{sin20^\circ +sin50^\circ +sin110^\circ }=[/tex]  

[tex]\bf =\dfrac{48\cdot \Big(\dfrac{1}{2}(sin110^\circ +sin50^\circ )-\dfrac{1}{2}(sin180^\circ -sin20^\circ )\Big)}{sin20^\circ +sin50^\circ +sin110^\circ }=[/tex]            

[tex]\bf =\dfrac{48\cdot \dfrac{1}{2}\cdot \Big(sin110^\circ +sin50^\circ -\overbrace{\bf sin180^\circ }^{0}+sin20^\circ \Big)}{sin20^\circ +sin50^\circ +sin110^\circ }=\\\\\\=\dfrac{24\cdot \Big(sin110^\circ +sin50^\circ +sin20^\circ \Big)}{sin20^\circ +sin50^\circ +sin110^\circ }=24[/tex]