NIMCET 2014 Question Paper with solutions

Let AB = x , OC = d , OM = h
Now , in ΔAOM
tan 30° = $\frac{\mathit{O}\mathit{M}}{\mathit{A}\mathit{O}}$
⇒ OA = $\sqrt{3}$ OM ... (i)
In ΔBOM
tan 45° = $\frac{\mathit{O}\mathit{M}}{60+\mathit{d}}$
⇒ h = 60 + d ... (ii)
In ΔMOC
tan 60° = $\frac{\mathit{O}\mathit{M}}{\mathit{O}\mathit{C}}$
⇒ $\sqrt{3}$ = $\frac{\mathit{h}}{\mathit{d}}$
⇒ $\sqrt{3}$ = $\frac{60+\mathit{d}}{\mathit{d}}$
d = 30 $\left(\sqrt{3}+1\right)$ ... (iii)
From Eq. (i) x + 60 + d = $\sqrt{3}$ h
⇒ x + 63 + 30 $\left(\sqrt{3}+1\right)$ = $\sqrt{3}$ (60 + 30 ($\sqrt{3}$ + 1))
⇒ x = $60\sqrt{3}$ + 90 + $30\sqrt{3}$ - 60 - $30\sqrt{3}$ - 30
⇒ x = 60 $\sqrt{3}$ ∴ AB = 60 $\sqrt{3}$