Compute area A given that A is a triangle inscribed in a unit circle with one vertex at $P = (1,0)$ and the other two at the points $P_1$ and $P_2$ given by polar coordinates with angles
$\theta_1$ and $\theta_2$, where $\theta_2 > \theta_1$.
My attempt at a solution:
I attempted to take the cross product of $\vec{PP_1}$ and $\vec{PP_2}$ by setting a $z$ coordinate for both vectors as $0$:
$\vec{B} =\vec{PP_1} \times \vec{PP_2} = (\sin(\theta_2)(1-\cos(\theta_1)) – \sin(\theta_1)(1-\cos(\theta_2)))\hat{k}$
$A = 0.5|\vec{B}| = 0.5(\sin(\theta_1 – \theta_2) + \sin(\theta_2) – \sin(\theta_1)$
Is there a more efficient way to do this problem? The polar coordinates could be converted to (x,y) to use Heron's Formula, but it does not seem to be the most optimal way.
Best Answer
This seem pretty efficient to me. You can get the same formula directly by noticing that $$A_{\triangle POP_1}+A_{\triangle P_1OP_2}=A_{\triangle POP_2}+A_{\triangle PP_2P_1}$$ On the unit circle, these areas are $\frac{1}{2}\sin\theta_1$, $\frac{1}{2}\sin(\theta_2-\theta_1)$, $\frac{1}{2}\sin\theta_2$, and $A$