Given polar co-ordinates are

(i) $$(1 , 30^{\circ})$$

Then, $$r = 1$$ and $$\theta = \dfrac{\pi}{6}$$

$$x = rcos{\theta}= 1\times cos\left(\dfrac{\pi}{6}\right) =\dfrac{\sqrt{3}}{2}$$

$$y = rcos{\theta}= 1\times sin\left(\dfrac{\pi}{6}\right) =\dfrac{1}{2}$$

$$A \left(\dfrac{\sqrt{3}}{2} , \dfrac{1}{2}\right) $$

(ii) $$(2 , 60^{\circ})$$

Then, $$r = 2$$ and $$\theta = \dfrac{\pi}{3}$$

$$x = rcos{\theta}= 2\times cos\left(\dfrac{\pi}{3}\right) =1$$

$$y = rcos{\theta}= 2\times sin\left(\dfrac{\pi}{3}\right) =\sqrt{3}$$

$$B (1 , \sqrt{3}) $$

(iii) $$(3 , 90^{\circ})$$

Then, $$r = 1$$ and $$\theta = \dfrac{\pi}{2}$$

$$x = rcos{\theta}= 3\times cos(\dfrac{\pi}{2}) =0$$

$$y = rcos{\theta}= 3\times sin(\dfrac{\pi}{2}) =3$$

$$C (0 , 3) $$

Now we have points $$A \left(\dfrac{\sqrt{3}}{2} , \dfrac{1}{2}\right) $$, $$B (1 , \sqrt{3}) $$ and $$C (0 , 3) $$

Then, area of triangle is given by the formula

$$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$$

$$A(\triangle ABC)=\dfrac{1}{2}\left | \dfrac{3}{2}+ 3+0-\dfrac{1}{2} -0-\dfrac{3\sqrt{3}}{2} \right | $$

$$\Rightarrow A(\triangle ABC)= \dfrac{8-3\sqrt{3}}{4} $$