question_answer

Each side of a square subtends an angle of\[{{60}^{o}}\]at the top of a tower\[h\]metre high standing in the centre of the square. If a is the length of each side of the square, then

A) \[2{{a}^{2}}={{h}^{2}}\]                

B) \[2{{h}^{2}}={{a}^{2}}\]

C) \[3{{a}^{2}}=2{{h}^{2}}\]             

D)        \[2{{h}^{2}}=3{{a}^{2}}\]

Correct Answer: B

Solution :

Let\[ABCD\]be a square of each side of length\[a\]. It is given that\[\angle OCP={{60}^{o}}\]. Length of diagonal                 \[AC=\sqrt{{{a}^{2}}+{{a}^{2}}}=a\sqrt{2}\] \[\therefore \]  \[PC=\frac{AC}{2}=\frac{a}{\sqrt{2}}\] In            \[\Delta \,\,OCP,\,\,\tan {{60}^{o}}=\frac{OP}{PC}\] \[\Rightarrow \]               \[\sqrt{3}=\frac{h}{a/\sqrt{2}}\] \[\Rightarrow \]               \[\sqrt{3}a=\sqrt{2}h\] \[\Rightarrow \]               \[3{{a}^{2}}=2{{h}^{2}}\]