Answer:
Let height AB of tower = h. In \[\Delta \,ABC\], \[\frac{AB}{BC}=\tan \,(90-\theta )\] \[\frac{h}{4}=\cot \,\theta \] ?(i) In \[\Delta \,ABD\], \[\frac{AB}{BC}=\tan \,\theta \] \[\frac{h}{16}=\tan \,\theta \] ?(ii) Multiply eq. (i) and (ii) \[\frac{h}{4}\times \frac{h}{16}=\cot \,\,\theta \,\times \tan \,\theta \] \[\frac{{{h}^{2}}}{64}=1\] [\[\because \] \[\cot \,\theta \times \tan \,\theta =\frac{1}{\tan \,\theta }\times \tan \,\theta =1\]] \[\Rightarrow \] \[{{h}^{2}}=64\Rightarrow h=8\,m\] Height of tower\[=8\text{ }m\].
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