• # 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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