Question

Find the areas of the triangles the coordinates of whose angular points are $(1,{30}^o),(2,{60}^o)$ and $(3,{90}^o)$

Answer 1

Given polar co-ordinates are

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

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

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

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

$A(\frac{\sqrt{3}}{2},\frac{1}{2})$

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

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

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

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

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

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

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

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

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

$C(0,3)$

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

Then, area of triangle is given by the formula

$△=\frac{1}{2}|x_1{y}_2+x_2{y}_3+x_3{y}_1-x_2{y}_1-x_3{y}_2-x_1{y}_3|$

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

$\Rightarrow A\left(△ABC\right)=\frac{8-3\sqrt{3}}{4}$