10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

  • question_answer 19) A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as \[60{}^\circ \] and the angle of depression of the base of hill as \[30{}^\circ \]. Find the distance of the hill from the ship and the height of the hill.

    Answer:

    Let AB be the height of water level and CD be the height of hill
    Then,
    In \[\Delta \,ABC\] \[\Delta \,ABC\]
                            \[\tan \,30{}^\circ =\frac{10}{y}\]
                                    \[y=10\sqrt{3}\]                                         ?(i)
    In \[\Delta \,ADE\]
                            \[\tan \,60{}^\circ =\frac{x}{y}\]
                                  \[y=\frac{x}{\sqrt{3}}\]                                           ?(ii)
                From (i) and (ii), we get
                                        \[\frac{x}{\sqrt{3}}=10\sqrt{3}\]
                                        \[x=10\times 3=30\,m\]
    \[\therefore \] Distance of the hill from this ship is \[10\sqrt{3}\,m\] and the height of the hill is \[30+10=40\text{ }m\].


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