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how do you find the arc length for $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{PQR}$?

I know $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{PQ}$ is $8.1$ do I just $8.1+8.1$ to find $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{PQR}$?

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    What exactly is 6.48 inches? – 2017-01-26
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    arc $PQR$ means the arc from $P$ to $R$ that passes through $Q$ (as opposed to arc $PSR$, which goes the other away round). – 2017-01-26
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    @suomynonA I'm confident it's the radius that is 6.48 inches. – 2017-01-26

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Arc length = Radius * Angle subtended by the arc at the centre(in radian)

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The circumference of the circle is $40.715$ inches. Sector PQ is $72^\circ$. That means that PQR is $\dfrac{137}{360}$ of the circumference. $40.715 \times \dfrac{137}{360}\approx15.4943$ inches.

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In a circle, we know that all the angles must have sum upto $360^\circ$. So for the angle we don't know we can write:

$$73^\circ+150^\circ+65^\circ+x^\circ=360^\circ\space\Longleftrightarrow\space x^\circ=360^\circ-\left(73^\circ+150^\circ+65^\circ\right)=72^\circ\tag1$$

Now, for the circumference of a circle we have:

$$\mathcal{S}=2\pi\cdot\text{r}\tag2$$

And, when we have a sector of the circle:

$$\mathcal{S}_\Delta=2\pi\cdot\frac{\theta^\circ}{360^\circ}\cdot\text{r}\tag3$$

So, in your example we have that:

$$\theta^\circ=x^\circ+65^\circ=72^\circ+65^\circ=137^\circ\tag4$$

So, the arc length will be:

$$\mathcal{S}_\text{PQR}=2\pi\cdot\frac{137^\circ}{360^\circ}\cdot6.48=\frac{1233\pi}{250}\approx15.4943349\space\text{inch}\tag5$$