In the figure below, two circular curves y and x create 60° and 90° angles with their respective centres. If the length of the bottom curve Y is 10$$\pi\ $$, find the length of the other curve.

Let P and Q be the centres of the circles with arcs x and y respectively.

Thus, $$\angle APB = 90$$ and $$\angle AQB = 60$$

Also, length of arc $$y = 10 \pi$$ cm

=> $$\frac{\theta}{360} \times 2 \pi r = 10 \pi$$

=> $$\frac{1}{6} \times 2 \times r = 10$$

=> $$r = AQ = 10 \times 3 = 30$$ cm

=> AB = 30 ($$\because \triangle$$ AQB is equilateral triangle)

Also, $$\triangle$$ APB is right isosceles triangle, => $$AP = \frac{30}{\sqrt{2}}$$ 

$$\therefore$$ Arc length = $$x = \frac{90}{360} \times 2 \pi \times \frac{30}{\sqrt{2}}$$

= $$\frac{15 \pi}{\sqrt{2}}$$