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

  • question_answer 26) The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is \[60{}^\circ \]. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is \[45{}^\circ \]. Find the height of the tower PQ and the distance PX (Use \[\sqrt{3}=1.73\])

    Answer:

    We have, PQ as a vertical tower
    In \[\Delta \,YZQ\]                       \[tan\text{ }45{}^\circ =\frac{QZ}{YZ}\]
                                              \[\frac{QZ}{YZ}=1\]
                                              \[QZ=YZ\]                           ?(i)
    And, in \[\Delta \text{ }XPQ\]              \[tan60{}^\circ =\frac{QP}{XP}\]
                                              \[\sqrt{3}=\frac{QZ+40}{XP}\]
                                              \[\sqrt{3}=\frac{QZ+40}{YZ}\]                    \[(\because \,XP=YZ)\]
                                       \[\sqrt{3}\,QZ=QZ+40\]              [Using (i)]
                               \[\sqrt{3}\,QZ-QZ=40\]
                              \[QZ\left( \sqrt{3}-1 \right)=40\]
                                        \[QZ=\frac{40}{\sqrt{3}-1}=\frac{40}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]
                                              \[=20\left( \sqrt{3}+1 \right)\]
                                              \[=20(2.73)\]
                                              \[=54.60\,m\]
                 \[\therefore PX=54.6\,m\]
    And \[PQ=(54.6+40)m=94.6\,m\]          


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