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Two office towers are 30 m apart. From the 15th floor (which is 40 m above the ground) of the shorter tower, the angle of elevation to the top of the other tower is 70 degrees. 

Determine the angle of depression to the base of the taller tower from the 15th floor.

Determine the height of the taller tower. 

Let $\varphi$ the angle of depression

$\begin{array}{|rcll|} \hline \tan{\varphi} &=& \frac{40\ m}{30\ m} \\ \tan{\varphi} &=& \frac{4}{3} \\ \varphi &=& 53.1301023542^{\circ} \\ \hline \end{array}$

The angle of depression to the base of the taller tower from the 15th floor is $53.1301^{\circ}$

Let H = height of the taller tower

$\begin{array}{|rcll|} \hline \tan{70^{\circ}} &=& \frac{H-40\ m}{30\ m} \\ 30\cdot \tan{70^{\circ}} &=& H-40 \\ H &=& 40 + 30\cdot \tan{70^{\circ}} \\ H &=& 40 + 30\cdot 2.74747741945 \\ H &=& 40 + 82.4243225836 \\ H &=& 122.424322584\ m \\ \hline \end{array}$

The height of the taller tower is 122.42 m