SSC CHSL Model Paper 15

Sum of interior angles of pentagon = (2n - 4) × 90°
= (2 × 5 - 4) × 90° = 6 × 90° = 540°
∴ ∠6 + ∠7 + ∠8 + ∠9 + ∠10 = 540°
Again,
∠1 + 180° - ∠7= 16
Exterior angle = Sum of two other interior angles
∠2 + ∠180°- ∠6 = ∠10
∠3 + ∠180°- ∠10 = ∠9
∠4 + ∠180°- ∠9 = ∠8
∠5 + ∠180°- ∠8 = ∠7
On adding,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 2 × 540 - 900° = 180°
∴ Required answer = 540° + 180° = 720°