SSC CGL 19 Oct 2014 Shift 2 Paper

∠ DAC = $\frac{1}{2}$ (180° – A)
= 90° – Δ ∠ $\frac{\mathit{A}}{2}$
In Δ ADB,
∠DAB + ∠ ADB + ∠DBA = 180°
⇒ 90° - ∠ $\frac{\mathit{A}}{2}$ + ∠ A + 2 ∠ B = 180°
[Since AD = AB ⇒ ∠ ADB = ∠ DBA = ∠ B]
⇒ 90° + ∠ $\frac{\mathit{A}}{2}$ + 2 ∠ B = 180°
⇒ ∠ A + 4 ∠ B = 180° ... (i)
In Δ ABE,
AB = BE
∴ ∠BAE = ∠ AEB
∴ ∠ ABE + ∠ BAE + ∠ BEA = 180°
90° - $\frac{\mathit{B}}{2}$ + ∠ B + 2 ∠ A = 180°
⇒ 90 ° + $\frac{\angle \mathit{B}}{2}$ + 2∠A = 80°
⇒ ∠ B + 4 ∠ A = 180° ....(ii)
From equations (i) and (ii),
∠ A + 4 ∠ B = 4 ∠ A + ∠ B
⇒ 3 ∠ A = 3 ∠ B
⇒ ∠ A = ∠ B
∴ ∠ A + 4 ∠ B = 180°
⇒ 5 ∠ B = 180°
⇒ ∠ B = $\frac{180}{5}$ = 36° - ∠ A
∴ ∠ ACB = 180° - 36° - 36°
= 180° - 72° = 108°