question_answer

The elevation of an object on a hill is observed from a certain point in the horizontal plane through its base, to be \[{{30}^{o}}.\] After walking 120 m towards it on level ground the elevation is found to be \[{{60}^{o}}.\] Then the height of the object (in metres) is

A)  120                       

B)         \[60\sqrt{3}\]      

C)         \[120\sqrt{3}\]   

D)         \[60\]

Correct Answer: B

Solution :

Let h be the height of the object. In \[\Delta CAD\]                                 \[\tan {{30}^{o}}=\frac{CD}{AC}\]                 \[\Rightarrow \]               \[\frac{1}{\sqrt{3}}=\frac{h}{120+x}\] \[\Rightarrow \]               \[\sqrt{3}h=120+x\]                        ?(i) and in \[\Delta CBD\] \[\tan {{60}^{o}}=\frac{CD}{BC}\]                             \[\Rightarrow \]               \[\sqrt{3}=\frac{h}{x}\]                 \[\Rightarrow \]               \[h=\sqrt{3}x\]                 ?(ii)                 From Eqs (i) and (ii),we get                                 \[3x=120+x\]                 \[\Rightarrow \]               \[x=60\,m\] On putting \[x=60\] in Eq. (i), we get Height of the object \[=60\sqrt{3}m\]