A circle is inscribed inside a sector of a circle. Given the radii of both , find the length of segment formed by joining the endpoints of the sector.

Question

$AOB$ is a sector of a circle with center $ O$ and radius $OA = 10$.
Circle with radius $3$ is inscribed in this sector such that it touches radius $OA$, radius $OB$ and arc $AB$.
Find the length of the chord $AB$.

I don't know where to begin. To calculate the length of $AB$ ,
we'll need the length of perpendicular from $O$ to $AB$.
( then we can use pythagoras theorem to get half of $AB$ and then $AB$) .
But how can I find that?

Answer 1

The distance from the centre of the inscribed circle to point $O$ is $7$, so the angle between $AO$ and the perpendicular line from $O$ to $AB$ is $\theta=\arcsin\left(\frac{3}{7}\right)$, assuming that lines $OA$ and $OB$ are tangents to the inscribed circle.

Thus the length of $AB$ is

$$AB=2\times OA\times\sin(\theta)=2\times10\times\frac{3}{7}=8\frac{4}{7}$$

Answer 2

Suppose that the center of the smaller circle is $O'$. Since $OA, OB$ are tangents to the smaller circle, then $O'O$ bisects $\angle AOB$. Let $\angle O'OB = \theta$, then $\sin \theta = \frac {3} {7} $, we can find $\cos 2\theta$ by the formula $\cos 2\theta = 1-2\sin^2 \theta$ and we get that $\cos 2\theta = \frac {31} {49}$. Now by Cosine Law we get: $$OA^2 +OB^2 -2\cdot OB\cdot OA\cdot \cos 2\theta =AB^2 $$ $$\Rightarrow 10^2 +10^2 -2\cdot 10^2\cdot \frac {31} {49} =AB^2 $$ $$\Rightarrow 200(1-\frac {31} {49})=AB^2 $$ $$AB=\frac {60} {7}=8\frac {4}{7}$$