A) \[817.8\text{ }m\]
B) \[818.5\text{ }m\]
C) \[820.5\text{ }m\]
D) \[819.6\text{ }m\]
Correct Answer: D
Solution :
Let \[BC=h\]be the cliff, A be the point of first observation and \[\angle CAB={{45}^{o}}\]. After covering \[AD=600\text{ }m,\] \[\angle DAE={{30}^{o}}.\] Draw \[DF\bot SC\]and \[DE\bot AB\] such that \[\angle CDF={{60}^{o}}\](Given) In \[\Delta DFC,\,\,\angle DCF={{180}^{o}}-({{90}^{o}}+{{60}^{o}})={{30}^{o}}\] In \[\Delta ABC,\,\,\angle ACB={{180}^{o}}-({{90}^{o}}+{{45}^{o}})={{45}^{o}}\] \[\therefore \] \[\angle ACD=\angle ACB-\angle DCF\] \[={{45}^{o}}-{{30}^{o}}={{15}^{o}}\] Also, \[\angle CAD={{45}^{o}}-{{30}^{o}}={{15}^{o}}\] In \[\Delta \text{ }ACD,\text{ }AD=DC\][Opposite sides of equal angles] \[\therefore \] \[DC=600\text{ }m\] \[[\because \,\,\,AD=600\,m]\] In right \[\Delta \,AED\], \[\sin {{30}^{o}}=\frac{DE}{AD}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{DE}{600}\,\,\Rightarrow \,2DE=600\,\,\,\Rightarrow \,\,DE=300\,m\] In rt \[\Delta \,CFD,\] \[\sin {{60}^{o}}=\frac{FC}{DC}\] \[=\frac{\sqrt{3}}{2}=\frac{FC}{600}\,\,\Rightarrow \,\,\,2FC=600\sqrt{3}\] \[\Rightarrow \] \[FC=300\sqrt{3}=519.6\,m\] Height of cliff \[BC=BF+FC=DE+FC\] \[(\because \,\,\,BF=DE)\] \[=300+519.6=819.6\text{ }m\].You need to login to perform this action.
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