If tan A = $\frac{1.1}{6}$, then what is the value of (4cos A - 7sin A) ? Given that A is an acute angle ?

$2\frac{41}{61}$

tanA = \(\frac{1.1}{6}\) = \(\frac{11}{60}\)

{ tan A = = \(\frac{P}{B}\) }

By using pythagoras theorem ,

P² + B² = H²

11² + 60² = H²

H = 61

Now,

4cos A - 7sin A

= 4 × \(\frac{B}{H}\)  - 7 × \(\frac{P}{H}\)

= 4 × \(\frac{60}{61}\)  - 7 × \(\frac{11}{61}\)

= 2 × \(\frac{41}{61}\)