(b) 132°
∠a = \(\frac{(5-2)\times180°}{5}=108°\) (Int. ∠ of a pentagon)
∠b = \(\frac{180°-180°}{2}=36°\) (∵ ΔABC is isos.)
∠c = \(\frac{(6-2)\times180°}{6}=120°\) (Int. ∠ of a hexagon)
∠e = 180° – 120° = 60° ( ∵SR || QC, co-int. ∠s)
∴ ∠x = ∠y – ∠b –∠e
= ∠BCD + ∠DCS – ∠b – ∠e
= 108° + 120° – 36° – 60°
= 132°