Question

Let $r_1$and $r_2$be positive real numbers and$0<{\theta }_2-{\theta }_1<\pi$. Prove that the area of the triangle with verticesrole="math" localid="1663155755932" $(0,0),(r_1,{\theta }_1),\mathrm{and}(r_2,{\theta }_2)$ in the polar plane is$\frac{r_1{r}_2}{2}\sin ({\theta }_2-{\theta }_1)$

Short answer

Hence, proved

Step-by-step solution

Given information

The vertices of the triangle are:

$(0,0),(r_1,{\theta }_1),\mathrm{and}(r_2,{\theta }_2)$

Area of the triangle

The area of the triangle is given by,

$A=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\mathrm{sq}.\mathrm{units}$

Here,

$$\begin{gathered} x_1=r_1\cos {\theta }_1,y_1=r_1\sin {\theta }_1 \\ {x}_2=r_2\cos {\theta }_2,y_2=r_2\sin {\theta }_2 \\ {x}_3=0,y_3=0 \end{gathered}$$

Now,

$$\begin{gathered} A=\frac{1}{2}\left[r_1\cos {\theta }_1\left(r_2\sin {\theta }_2-0\right)+r_2\cos {\theta }_2\left(0-r_1\sin {\theta }_1\right)+0\left(r_1\sin {\theta }_1-r_2\sin {\theta }_2\right)\right]\mathrm{sq}.\mathrm{units} \\ A=\frac{1}{2}\left[r_1{r}_2\cos {\theta }_1\sin {\theta }_2-r_2{r}_1\sin {\theta }_1\cos {\theta }_2\right]\mathrm{sq}.\mathrm{units} \\ A=\frac{1}{2}\left[r_1{r}_2\left(\cos {\theta }_1\sin {\theta }_2-\sin {\theta }_1\cos {\theta }_2\right)\right]\mathrm{sq}.\mathrm{units} \\ A=\frac{1}{2}\left[r_1{r}_2\left(\sin \left({\theta }_2-{\theta }_1\right)\right)\right]\mathrm{sq}.\mathrm{units} \\ A=\frac{r_1{r}_2\sin \left({\theta }_2-{\theta }_1\right)}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$

Hence, proved.