JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are \[{{\cot }^{-1}}\left( 3\sqrt{2} \right)\] and \[\cos e{{c}^{-1}}\left( 2\sqrt{2} \right)\] respectively,  then the height of the tower (in metres) is : [JEE Main 10-4-2019 Morning]

    A) \[10\sqrt{5}\]              

    B) \[\frac{100}{3\sqrt{3}}\]

    C) \[20\]   

    D) \[25\]

    Correct Answer: C

    Solution :

    \[\cot \alpha =3\sqrt{2}\]\[\And cosec\beta =2\sqrt{2}\]                     So,\[\frac{x}{h}=3\sqrt{2}\]                                                ...(i) And \[\frac{h}{\sqrt{{{10}^{4}}-{{x}^{2}}}}=\frac{1}{\sqrt{7}}\]                             ?(ii) So, from (i) & (ii) \[\Rightarrow \frac{h}{\sqrt{{{10}^{4}}-18{{h}^{2}}}}=\frac{1}{\sqrt{7}}\]\[\Rightarrow 25{{h}^{2}}=100\times 100\] \[\Rightarrow h=20.\]             


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