JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Height and Distance

  • question_answer
    The angular elevation of a tower CD at a point A due south of it is\[{{60}^{o}}\]and at a point B due west of A, the elevation is\[{{30}^{o}}\]. If AB = 3 km, the height of the tower is [MP PET 1998]

    A) \[2\sqrt{3}\,km\]

    B) \[2\sqrt{6}\,km\]

    C) \[\frac{3\sqrt{3}}{2}km\]

    D) \[\frac{3\sqrt{6}}{4}km\]

    Correct Answer: D

    Solution :

    From \[\Delta \text{ }CDA,\,x=h\cot {{60}^{o}}\]=\[\frac{h}{\sqrt{3}}\] From \[\Delta \,CDB,\,y=h\cot {{30}^{o}}=\sqrt{3}h\] From \[\Delta ABC\], by Pythagoras theorem, \[{{x}^{2}}+{{3}^{2}}={{y}^{2}}\] \[{{c}_{2}}\]  \[\left( \frac{h}{\sqrt{3}} \right)+{{3}^{2}}={{(\sqrt{3}h)}^{2}}\Rightarrow h=\frac{3\sqrt{6}}{4}\] km.

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