question_answer

A man on the top of a cliff 100 m high, observe the angles of depression of two points on the opposite sides of the cliff as \[\text{3}{{\text{0}}^{\text{o}}}\] and \[\text{6}{{\text{0}}^{\text{o}}}\] respectively. Then, the distance between the two points is equal to

A) \[400\sqrt{3}\,m\]          

B)         \[\frac{400}{\sqrt{3}}\,m\]        

C)         \[\frac{100}{\sqrt{3}}\,m\]        

D)         \[200\sqrt{3}\,m\]

Correct Answer: B

Solution :

Let PQ be the cliff and A and B be the points under Observations                 \[\therefore \]  \[PQ=100\,m\] In right angled \[\Delta \,APQ,\]                                 \[\frac{AQ}{PQ}=\cot {{30}^{o}}\]                 \[\Rightarrow \]               \[AQ=PQ.\sqrt{3}\]                 \[\Rightarrow \]               \[AQ=100\sqrt{3}\]                         ?(i) In right angled \[\Delta BPQ,\]                                 \[\frac{BQ}{PQ}=\cot {{60}^{o}}\]                 \[\Rightarrow \]               \[BQ=\frac{100}{\sqrt{3}}\]                         ?(ii) Adding Eqs. (i) and (ii), we get                 \[AQ+BQ=100\left( \sqrt{3}+\frac{1}{\sqrt{3}} \right)\]                 \[\Rightarrow \]               \[AB=100\frac{(3+1)}{\sqrt{3}}=\frac{400}{\sqrt{3}}m\]