Question

see photo plz to solve it​

Answer 1

Step-by-step explanation:

$\tan( \beta ) = \frac{4}{x} \\ \\ \tan( \alpha + \beta ) = \frac{7}{x} \\ \\ \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) } = \frac{7}{x} \\ \\ \frac{ \tan( \alpha ) + \frac{4}{x} }{1 - ( \frac{4}{x}) \tan( \alpha ) } = \frac{7}{x} \\ \\ x \tan( \alpha ) + 4 = 7 - \frac{28}{x} \tan( \alpha ) \\ \\ {x}^{2} \tan( \alpha ) + 4x = 7x - 28\tan( \alpha ) \\ \\ ( {x}^{2} + 12) \tan( \alpha ) = 7x - 4x \\ \\ \tan( \alpha ) = \frac{ 3 x}{ {x}^{2} + 12} \: .$