10th Class Mathematics Solved Paper - Mathematics-2016 Delhi Term-II Set-I

  • question_answer The angles of depression of the top and bottom of a 50 m high building from the top of a tower are \[45{}^\circ \] and \[60{}^\circ \] respectively. Find the height of the tower and the horizontal distance between the tower and the building, (use \[\sqrt{3}=1.73\])

    Answer:

    Let AB and CD be the tower and high building respectively
    Given,               \[CD=50\text{ }m\]
    Let,                   \[AB=h\text{ }m\]
    Then, in \[\Delta \,ADE\]
                            \[\tan \,45{}^\circ =\frac{AE}{DE}\]
                                     \[1=\frac{h-50}{DE}\]
                                 \[DE=h-50\]                                      ?(i)
    and, in \[\Delta \text{ }ACB\]
                            \[\tan \,60{}^\circ =\frac{AB}{CB}\]
                                  \[\sqrt{3}=\frac{h}{CB}\]
                                 \[CB=\frac{h}{\sqrt{3}}\]                                          ?(ii)
    Now, \[CB=DE\]
    then from eq. (i) and (ii), we get
                            \[h-50=\frac{h}{\sqrt{3}}\]
                            \[h-\frac{h}{\sqrt{3}}=50\]
                       \[\frac{\left( \sqrt{3}-1 \right)}{\sqrt{3}}h=50\]
                                   \[h=\frac{50\sqrt{3}}{\sqrt{3}-1}=\frac{50\sqrt{3}}{\sqrt{3}-1}\times \frac{50\sqrt{3}}{\sqrt{3}-1}\]
                                      \[=25\times 3+25\sqrt{3}\]
                                   \[h=75+25(1.73)\]
                                      \[=118.25\,m\]
    Hence, the height of the tower is \[118.25\text{ }m\]and the horizontal distance between the tower and the building is\[68.25\text{ }m\].


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