Answer:
Let C be top of a \[7m\] building CD and AB be tower. From C, draw \[CE\bot AB\], so EBDC is a rectangle. From \[\Delta \text{ }CBD,\text{ }tan\text{ }45{}^\circ =\frac{CD}{BD}\] or \[BD=CD=7\,m\] From \[\Delta \,AEC\] \[\frac{AE}{EC}=\tan \,\,60{}^\circ \] \[\Rightarrow \] \[AE=EC\,\,\tan \,\,60=7\sqrt{3}\] \[[\because \,EC=BD]\] Height of tower is \[AB=AE+EB=AE+DC\] \[=7\sqrt{3}+7\] \[7\left( \sqrt{3}+1 \right)m\]
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