Sin x tends to x as x tends to zero

Question

If ABC is a sector of a circle with centre A, why does the area of triangle ABC approach the area of sector ABC as angle BAC approaches zero?

Answer 1

Note that if the angle at the centre is $\theta$ measured in radians, and the radius is $r$ then the area of the sector is $\cfrac {r^2\theta}2$.

The area of the triangle is calculated by taking $D$ as the midpoint of $BC$ so that $AD=r\cos\frac{\theta}2$ and $BC=2r\sin\frac{\theta}2$. The area is then $\frac 12 AD\cdot BC=\frac{r^2\sin \theta}2$.

Extending $AD$ to meet the circle at $E$ and taking the tangent at $E$ to meet $AB$ at $F$ and $AC$ at $G$, $AFG$ is similar to $ABC$, with the sides longer by a factor of $\frac 1{\cos {\frac{\theta}2}}$ so we get $$\frac{r^2\sin \theta}2\lt\cfrac {r^2\theta}2\lt\frac 1{\cos^2 {\frac{\theta}2}}\cdot\frac{r^2\sin \theta}2$$

Since $\frac 1{\cos^2 {\frac{\theta}2}}$ is approximately $1$ for small values of $\theta$, the two outside expressions become close to one another - you can make this as precise as you need.