Question

if 6 tan A -5 =0
find:
3 sin A minus cos A / 5 cos A+ 9 Sin A​

Answer 1

Answer:

3/25

Step-by-step explanation:

Given

6tan A - 5 = 0

tan A = 5/6

Now coming back to expression

$\sf = \dfrac{3sinA - cosA}{5cosA + 9sinA}$

Multiplying both numerator and denominator by cos A

$\sf = \dfrac{3sinA - cosA}{5cosA + 9sinA} \times \dfrac{cosA}{cosA}$

$\sf = \dfrac{3sinA - cosA}{cosA } \times \dfrac{cosA}{5cosA + 9sinA}$

$\sf = \dfrac{3sinA - cosA}{cosA } \div \dfrac{5cosA + 9sinA}{cosA}$

$\sf = \bigg( \dfrac{3sinA }{cosA } - \dfrac{cosA}{cosA} \bigg) \div \bigg( \dfrac{5cosA}{cosA} + \dfrac{9sinA}{cosA} \bigg)$

$\sf = \dfrac{ 3tanA - 1}{ 5+ 9tanA}$

$\sf = \dfrac{ 3 \bigg( \dfrac{5}{6} \bigg) - 1}{ 5+ 9 \bigg( \dfrac{5}{6} \bigg) }$

$\sf = \dfrac{ \dfrac{5}{2} - 1}{ 5+ \dfrac{15}{2} }$

$\sf = \dfrac{ \dfrac{5 - 2}{2} }{ \dfrac{10 + 15}{2} }$

$\sf = \dfrac{ \dfrac{3}{2} }{ \dfrac{25}{2} }$

$\sf = \dfrac{3}{25}$