Answer:
Let AB be hill and DC be tower. From\[\Delta \text{ }ABC\] \[\frac{AB}{BC}=\tan \,\,60{}^\circ \] \[h=BC\,\,\tan \,\,60{}^\circ =\sqrt{3}\,\,BC\] From \[\Delta \,DBC,\frac{DC}{BC}=\tan \,\,30{}^\circ =\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[BC=\sqrt{3}\,\,DC=50\sqrt{3}\] \[h=BC\sqrt{3}\] \[=50\sqrt{3}\times \sqrt{3}=50\times 3\] \[=150\,\,m\]
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