If tan A = $\frac{1.1}{6}$, then what is the value of (4cos A - 7sin A) ? Given that A is an acute angle ?
Answer & explanation
Correct answer: option 4
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 × \(\frac{41}{61}\)