question_answer

The angle of elevation of a cliff from a fixed point A is \[{{45}^{o}}\]. After going up a distance of 600 metres towards the top of the cliff at an inclination of \[{{30}^{o}},\] it is found that the angle of elevation is \[{{60}^{o}}\]. Find the height of the cliff.

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 \[BC=BF+FC=DE+FC\]                                      \[(\because \,\,\,BF=DE)\] \[=300+519.6=819.6\text{ }m\].