RRB NTPC Mains or Stage 2 Model Paper 4 with solutions

(b) Let the height of the tower be h m and∠ CBD = θ then ∠ DAC = 90º – θ
(Because both angles are complementary)

∴ In ΔBCD,
$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }=\frac{\mathit{C}\mathit{D}}{\mathit{B}\mathit{C}}\Rightarrow \mathit{t}\mathit{a}\mathit{n}\mathit{\theta }=\frac{\mathit{h}}{3}$
Now, in ΔACD
$\mathit{t}\mathit{a}\mathit{n}\left(90\mathit{º}\mathit{–}\mathit{\theta }\right)=\frac{\mathit{C}\mathit{D}}{\mathit{A}\mathit{C}}\Rightarrow \mathit{c}\mathit{o}\mathit{t}\mathit{\theta }=\frac{\mathit{h}}{12}$
$\frac{1}{\mathit{t}}\mathit{a}\mathit{n}\mathit{\theta }=\frac{\mathit{h}}{12}$
h tan θ =12
Put the value of tanθ
$\mathit{h}\times \frac{\mathit{h}}{3}=12$
${\mathit{h}}^2=36\therefore \mathit{h}=6$
Then, height of tower = 6 m.