If $$\theta$$ lies in the first quadrant and $$5 \tan \theta = 4$$, then $$\frac{5 \sin \theta - 3 \cos \theta}{\sin \theta + 2 \cos \theta} =$$

Solution 

Given $$5 \tan \theta = 4$$

So $$ tan \theta $$ = (4/5)

P = 4

B = 5

H=  $$\sqrt{\ \left(4^2\right)\ +\left(5^2\right)}$$ = $$\sqrt{41\ }$$

$$ sin \theta $$ = (4/$$\sqrt{41\ }$$)

$$ cos \theta $$ = (5/$$\sqrt{41\ }$$)

So 

= {5 (4/$$\sqrt{41\ }$$) - 3 (5/$$\sqrt{41\ }$$) } / {(4/$$\sqrt{41\ }$$) + 2 (5/$$\sqrt{41\ }$$) }

={20 $$\sqrt{41\ }$$ - 15 $$\sqrt{41\ }$$} / {4 $$\sqrt{41\ }$$ + 10 $$\sqrt{41\ }$$}

= $$\frac{5}{14}$$ Answer