Question

(7sin a +9 cos a ) / (5sin a -5cos a ), если tg a =6

Answer 1

$\displaystyle\bf\\tg\alpha =6\\\\\\\frac{Sin\alpha }{Cos\alpha } =6 \ \ \ \Rightarrow \ \ \ Sin\alpha =6Cos\alpha \\\\\\\frac{7Sin\alpha +9Cos\alpha }{5Sin\alpha -5Cos\alpha }=\frac{7\cdot 6Cos\alpha +9Cos\alpha }{5\cdot 6Cos\alpha -5Cos\alpha } =\frac{42Cos\alpha +9Cos\alpha }{30Cos\alpha -5Cos\alpha } =\\\\\\=\frac{51Cos\alpha }{25Cos\alpha } =\frac{51}{25} =2,04$