Elevation of the tower

We can see the top of the tower standing on a plane from a certain point A at an elevation angle of 39°25''. If we come towards its foot 50m closer to place B, we can see the top of the tower from it at an elevation angle of 56°42''. How tall is the tower?

Correct answer:

h =  89.3 m

Step-by-step explanation:

$$\begin{gathered} \alpha =56°42^{\prime }=56°+\frac{42}{60}°=56.7°=56.7 \\ \beta =39°25^{\prime }=39°+\frac{25}{60}°=39.4167°\doteq 39.4167 \\ a=50\text{m} \\ \tan \alpha =h/x \\ \tan \beta =h/(x+a) \\ x=h/\tan \alpha =h/t_1 \\ {t}_1=\tan \alpha =\tan 56.7°=1.522355=1.52235 \\ {t}_2=\tan \beta =\tan 39.416666666667°=0.821897=0.8219 \\ {t}_2\cdot (h/t_1+a)=h \\ {t}_2\cdot h/t_1+a\cdot t_2=h \\ h=\frac{a\cdot t_1\cdot t_2}{t_1-t_2}=\frac{50\cdot 1.5224\cdot 0.8219}{1.5224-0.8219}\doteq 89.3143\text{m} \\ \text{ Verifying Solution: } \\ x=h/t_1=89.3143/1.5224\doteq 58.6685\text{m} \\ {\alpha }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x})=\frac{180°}{\pi }\cdot \arctan (\frac{89.3143}{58.6685})=\frac{567}{10}=56.7{}^{\circ } \\ {\beta }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x+a})=\frac{180°}{\pi }\cdot \arctan (\frac{89.3143}{58.6685+50})=\frac{473}{12}\doteq 39.4167{}^{\circ } \end{gathered}$$

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