The area of the triangle formed by the lines $x=0;y=0$ and $x\sin { { 18 }^{ 0 } } +y\cos { { 36 }^{ 0 } } +1=0$ is-

$\begin{array}{l} \text { Area }=\frac{1}{2} \times A B \times A C \\ \Rightarrow \frac{1}{2} \times \frac{1}{\cos 36^{\circ}} \times \frac{1}{\sin 18^{\circ}} \\ \Rightarrow \frac{1}{2} \times \frac{\cos 18^{\circ}}{\cos 36^{\circ}} \times \frac{1}{\sin 18^{\circ} \cos 18^{\circ}} \\ \text { using } \sin 2 \theta=2 \sin \theta \cos \theta\end{array}$

$\begin{aligned} \Rightarrow \text { Area } & =\frac{\cos 18^{\circ}}{\cos 36^{\circ} \times \sin 36^{\circ}} \times \frac{2}{2} \\ & =\frac{2 \cos 18^{\circ}}{\sin 72^{\circ}} \\ \sin 72^{\circ} & =\cos \left(90^{\circ}-18^{\circ} \right)=\cos 18^{\circ} \\ \text { Area } & =\frac{2 \cos 18^{\circ}}{\cos 18^{\circ}} \\ \text { ∴ Area } & =2\end{aligned}$