Find a formula for the area of the triangle with vertices (in polar coordinates) $(r_1,\theta_1)$, $(r_2, \theta_2)$, and $(r_3,\theta_3)$.

I don't know where to start with this homework problem. If anyone could help me start, or provide the solution, that'd be greatly appreciated.

1 Answer

You don't need to convert to Cartesian coordinates.

Let the triangle vertices be $A,B$ and $C$ with the polar coordinates as shown below.

Then use the following formula for the area of a triangle whose side lengths are $a$ and $b$ and the included angle between them is $\theta$$$\text{area}=\frac{1}{2}ab \sin \theta .$$

With this

1. the area of $\triangle ADB=\frac{1}{2}(|AD|\,|BD|) \sin (\angle ADB)=\frac{1}{2}|r_1r_2| \sin\left(\theta_2-\theta_1\right)$.
2. the area of $\triangle BDC=\frac{1}{2}|r_2r_3| \sin\left(\theta_3-\theta_2\right)$.
3. the area of $\triangle ADC=\frac{1}{2}|r_1r_3| \sin\left(\theta_3-\theta_1\right)$.

Now the area of $\triangle ABC=\triangle ADB+\triangle BDC-\triangle ADC$. So$$\text{Area of } \triangle ABC=\color{magenta}{\frac{1}{2}\left[|r_1r_2| \sin\left(\theta_2-\theta_1\right)+|r_2r_3| \sin\left(\theta_3-\theta_2\right)-|r_1r_3| \sin\left(\theta_3-\theta_1\right)\right]}$$