If 5 tan A = 12, then what is the value of $\frac{13sinA+20tanA}{15tanA-13cosA}$, where A is an acute angle ?
Answer & explanation
Correct answer: option 4
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\(\frac{29}{ 31}\)