# 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.

$$AOB$$ 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?

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
$$AB=2\times OA\times\sin(\theta)=2\times10\times\frac{3}{7}=8\frac{4}{7}$$
• Glad you like the solution. The arcsin is useful when you can find a right angled triangle with two sides you can easily calculate - which was the triangle formed by the centre of the inscribed circle, point $O$ and the point where $OA$ touches the inscribed circle. Once you have found the angle, you can easily work out $AB$. – Alijah Ahmed Apr 4 '14 at 18:01
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: $$OA^2 +OB^2 -2\cdot OB\cdot OA\cdot \cos 2\theta =AB^2$$ $$\Rightarrow 10^2 +10^2 -2\cdot 10^2\cdot \frac {31} {49} =AB^2$$ $$\Rightarrow 200(1-\frac {31} {49})=AB^2$$ $$AB=\frac {60} {7}=8\frac {4}{7}$$