An arc of length 15 cm subtends an angle of 45°

We know that the arc length l of a sector of an angle θ in a circle of radius r is

$l=\frac{\theta}{360^{\circ}} \times 2 \pi r$

It is given that $l=15 \mathrm{~cm}$ and angle $\theta=45^{\circ}$.

Now we substitute the value of l and θ in above formula to find the value of radius r of circle.

$15 \mathrm{~cm}=\frac{45^{\circ}}{360^{\circ}} \times 2 \pi r$

$r=\frac{15 \times 360^{\circ}}{2 \pi \times 45^{\circ}} \mathrm{cm}$

$r=\frac{60}{\pi} \mathrm{cm}$