Symbiosis National Aptitude Test SNAP 2017 Paper

2 sin α + 15 $\mathit{c}\mathit{o}{\mathit{s}}^2$ α = 7
⇒ 2 sin α + 15 (1 - $\mathit{s}\mathit{i}{\mathit{n}}^2$ α) = 7
⇒ 2 sin α + 1 5 - 15 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α = 7
⇒ 15 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α - 2 sin α - 8 = 0
⇒ (3 sin α + 2) (5 sin α - 4) = 0
sin α = -2/3 or 5 sin α - 4 = 0
sin α = $\frac{4}{5}$
Since, α lies between 0° - 90°. So, sin α is positive.
sin α = $\frac{4}{5}$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}$
Base = $\sqrt{{\left(5\right)}^2-{\left(4\right)}^2}$ = 3
∴ cot α = $\frac{\mathrm{Base}}{\mathrm{Perpendicular}}$ = $\frac{3}{4}$