SSC CGL 31 Aug 2016 Paper

5 cos θ + 12 sin θ = 13
⇒ 5 cos θ – 13 = 12 sin θ
On squaring both sides,
25 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + 169 – 130 cos θ
= 144 (1 – $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ)
⇒ 25 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ – 130 cos θ + 169
= 144 – 144 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ
⇒ 144 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + 25 $\mathit{c}\mathit{o}{\mathit{s}}^2$θ – 130
cos θ + 169 – 144 = 0
⇒ 169 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ – 130 cos θ + 25 = 0
⇒ ${\left(13\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{–}5\right)}^2$ = 0
⇒ 13 cos θ – 5 = 0
⇒ 13 cos θ = 5
OR
⇒ cosθ = $\frac{5}{13}$
5 cos θ + 12 θ = 13
⇒ $\frac{5}{13}$ cos θ + $\frac{13}{13}$ sin t = 1
Since $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }+\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ = 1
∴ cos θ = $\frac{5}{13}$