Finding length of arc intercepted by two tangent lines given radius and distance from circle

I'm in the process of writing a script for work but realized that I need to calculate the length of a circular arc intersected by two tangent lines, and when I try finding it I get lost in a mess of trig that seems to go nowhere.

Here's what I have so far:

Tangent lines intersect each other at point $$P$$

Tangent lines intersect circle at points $$A,B$$

Circle radius is $$R$$

Point on circle nearest to $$P$$ is $$Q$$

Length of $$\vec{PQ}$$ is $$T$$

Using some trig, I found that $$\vec{PA}=\vec{PB} = R\arccos(\frac{R}{R+T})$$

But that's where I'm stuck. How can I find $$\widehat{AQB}$$ ?

• Let the centre of the circle be $O$. The $\angle AOP = \angle POB = \arccos \left( \dfrac{R}{R + T} \right)$. The required arc length is $R \angle AOB = 2R \angle AOP = 2R \arccos \left( \dfrac{R}{R + T} \right)$. – sudeep5221 Feb 3 at 18:30
• @sudeep5221 This is great and a lot simpler than I imagined! One question though, why isn't the denominator in the arccos R+H? EDIT: Nevermind, I see that I made a typo in my original question. You are correct. – David Feb 3 at 18:39
• @sudeep5221 I think you should make this an answer. – saulspatz Feb 3 at 18:47

$$OAP$$ is a right triangle because $$PA$$ is tangent to the circle. With other given data we are able to find angle $$\widehat O$$ in this triangle: $$\cos O = \frac{OA}{OP} = \frac{R}{R+T}$$ $$\widehat {AOQ} = arccos \frac{R}{R+T}$$ Now, it is easy to show that $$\widehat{QOB}=\widehat{AOQ}$$ and therefore $$\widehat {AOB} = 2 arccos \frac{R}{R+T}$$ The above is the angle of the arc $$AQB$$. If you are looking for the other arc that connects $$A$$ to $$B$$, it is $$2 \pi - 2 arccos \frac{R}{R+T}$$. And if you are looking for the angle $$\widehat{AQB}$$ , it is half of the latter arc: $$\widehat{AQB} = \pi - arccos \frac{R}{R+T}$$ .
From $$\Delta AOP$$ subtended angle is directly (since $$OP$$ is hypotenuse)
$$2 \cos^{-1}\dfrac{R}{R+T}$$
$$R$$ times the above is the length of the arc subtended by this angle.