please solve the tenth

Dear student

G i v e n : \frac{a r c A P B}{2} = \frac{a r c B Q C}{3} = \frac{a r c C R A}{4} = k (s a y) a r c A P B = 2 k, a r c B Q C = 3 k a n d a r c C R A = 4 k A l s o, C i r c u m f e r e n c e o f c i r c l e = a r c A P B + a r c B Q C + a r c C R A = 2 k + 3 k + 4 k = 9 k A l s o, C i r c u f e r e n c e o f c i r l c e = 2 πr \Rightarrow 2 πr = 9 k N o w, L e n g t h o f a n a r c o f a s e c t o r o f ∠ B O C = \frac{∠ B O C}{360 °} \times 2 πr 3 k = \frac{∠ BOC}{360 °} \times 9 k ∠ BOC = 120 °

Regards