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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