An observer

An observer standing west of the tower sees its top at an altitude angle of 45 degrees. After moving 50 meters to the south, he sees its top at an altitude angle of 30 degrees. How tall is the tower?

Correct answer:

y =  35.3553 m

Step-by-step explanation:

$$\begin{gathered} A=45{}^{\circ } \\ B=30{}^{\circ } \\ {x}_3=50\text{m} \\ \tan A=y:x_1 \\ \tan B=y:x_2 \\ {x}_2^2=x_3^2+x_1^2=50^2+35.3553^2=2500 \\ y=x_1\cdot \tan A=x_2\cdot \tan B \\ {x}_1\cdot \tan A=x_2\cdot \tan B \\ {x}_1^2\cdot {\tan }^{2}A=x_2^2\cdot {\tan }^{2}B \\ {x}_1^2\cdot {\tan }^{2}A=(x_3^2+x_1^2)\cdot {\tan }^{2}B \\ {t}_1=(\tan A{)}^2=(\tan 45°{)}^2=1^2=1 \\ {t}_2=(\tan B{)}^2=(\tan 30°{)}^2=0.57735^2=0.33333 \\ {x}_1^2\cdot t_1=x_3^2\cdot t_2+x_1^2\cdot t_2 \\ {x}_1^2\cdot (t_1-t_2)=x_3^2\cdot t_2 \\ {x}_1=x_3\cdot \sqrt{\frac{t_2}{t_1-t_2}}=50\cdot \sqrt{\frac{0.3333}{1-0.3333}}=25\sqrt{2}\text{m}\doteq 35.3553\text{m} \\ y=x_1\cdot \sqrt{t_1}=35.3553\cdot \sqrt{1}=25\sqrt{2}\doteq 35.3553\text{m} \\ \text{ Verifying Solution: } \\ {x}_2=y/\sqrt{t_2}=35.3553/\sqrt{0.3333}=25\sqrt{6}\text{m}\doteq 61.2372\text{m} \\ {x}_{22}=\sqrt{x_3^2+x_1^2}=\sqrt{50^2+35.3553^2}=25\sqrt{6}\text{m}\doteq 61.2372\text{m} \end{gathered}$$

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