SSC CGL 2024 PYQ — Question 71

The correct answer is Option 4 i.e. $5\over 13$.

$2\ tan\ θ$ = 3

⇒ tan θ = 3/2 = P/B

By Pythagoras theorem,

⇒ H² = B² + P²

here, H  = Hypotaneous, B = Base, P = Perpendicular

⇒ H² = 2² + 3²

⇒ H² = 4 + 9

⇒ H² = 13

⇒ H = $\sqrt{13}$

So, sinθ = P/H = 3/$\sqrt{13}$

⇒ cosθ = B/H = 2/$\sqrt{13}$

Now,

⇒ $3\ sin\ θ\ - 2\ cos\ θ \over 3\sin\ θ\ +\ 2\ cos\ θ$

⇒ $\frac{{{3}\times {3\over{\sqrt13}}} - {{2}\times {2\over{\sqrt13}}}}{{{3}\times {3\over{\sqrt13}}}+ {{2}\times {2\over{\sqrt13}}}}$

⇒ $\frac{{9\over{\sqrt13}} - {4\over{\sqrt13}}}{{9\over{\sqrt13}}+{4\over{\sqrt13}}}$

⇒ $\frac{5\over{\sqrt13}}{13\over{\sqrt13}}$ = $5\over 13$