The angle 9

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 m towards the foot of the tower to a point B, the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from A .

Correct answer:

h =  17.3 m
x =  10 m

Step-by-step explanation:

$$\begin{gathered} \alpha =30{}^{\circ } \\ \beta =60{}^{\circ } \\ a=20\text{m} \\ \tan \beta =h:x=h/x \\ \tan \alpha =h:(x+a)=h/(x+a) \\ x=h/\tan \beta =h/t_1 \\ {t}_1=\tan \beta =\tan 60°=1.732051=1.73205 \\ {t}_2=\tan \alpha =\tan 30°=0.57735 \\ {t}_2\cdot (h/t_1+a)=h \\ {t}_2\cdot h/t_1+a\cdot t_2=h \\ h=\frac{a\cdot t_1\cdot t_2}{t_1-t_2}=\frac{20\cdot 1.7321\cdot 0.5774}{1.7321-0.5774}=10\sqrt{3}\doteq 17.3205\text{m} \\ \text{ Verifying Solution: } \\ x=h/t_1=17.3205/1.7321=10\text{m} \\ {\alpha }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x+a})=\frac{180°}{\pi }\cdot \arctan (\frac{17.3205}{10+20})=30{}^{\circ } \\ {\beta }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x})=\frac{180°}{\pi }\cdot \arctan (\frac{17.3205}{10})=60{}^{\circ } \end{gathered}$$

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