If 5 tan A = 12, then what is the value of $\frac{13sinA+20tanA}{15tanA-13cosA}$, where A is an acute angle ?

$1\frac{29}{31}$

5tanA = 12

tanA =  \(\frac{12 }{5}\)

{ tanA =  \(\frac{P}{B}\) }

P² + B² = H²

12² + 5² = H²

H = 13

Now,

  \(\frac{13 sinA + 20 tanA}{ 15tanA - 13 cosA}\)

= \(\frac{13× 12/13 + 20 ×12/5}{ 15×12/5 - 13 ×5/13}\)

= \(\frac{12+ 48}{ 36 - 5}\)

= \(\frac{60}{ 31}\)

= 1\(\frac{29}{ 31}\)