+1067 Points $L$ and $M$ lie on a circle $\omega_1$ centered at $O$. The circle $\omega_2$ passing through points $O,$ $L,$ and $M$ is drawn. If the measure of arc $LM$ in circle $\omega_1$ is $90^\circ,$ and the radius of $\omega_1$ is 1, then find the area of triangle $LOM$.
+98 The area of triangle LOM is just \((1/2) r^2 sin (90°)\).
\(sin(90) = 1\) amd the circle has radius 1, so we just have \((1/2)(1)^2 * (1) \)
This means our final answer is just \([ LOM ] = (1/2) r^2 sin (90°) = (1/2)(1)^2 * (1) = 1/2\).
1/2
Thanks! :)
