Answer:
Let AB be the tower and CD be a building of height 30 m and x m respectively. Let the distance between the two be y m, Then, in \[\Delta \text{ }ABC\] \[\frac{30}{y}=\tan \,45{}^\circ \] \[\frac{30}{y}=1\Rightarrow y=30\] And, in \[\Delta \,BDC\] \[\frac{x}{y}=\tan \,\,30{}^\circ \] \[x=y\,\,\tan \,\,30{}^\circ \] \[x=30\times \frac{1}{\sqrt{3}}=10\sqrt{3}\] Hence, the height of the building is \[10\sqrt{3}\,m\].
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