Angle of elevation 3

The angle of elevation of a pole from a point on the horizontal ground is 15°. After going up a distance of 10m towards the pole the angle of elevation became 30°.What is the height of the pole?

Correct answer:

h =  5 m

Step-by-step explanation:

$$\begin{gathered} \alpha =15{}^{\circ } \\ \beta =30{}^{\circ } \\ a=10\text{m} \\ \tan \beta =h:x=h/x \\ \tan \alpha =h:(x+a)=h/(x+a) \\ x=h/\tan \beta =h/t_1 \\ {t}_1=\tan \beta =\tan 30°=0.57735 \\ {t}_2=\tan \alpha =\tan 15°=0.267949=0.26795 \\ {t}_2\cdot (h/t_1+a)=h \\ {t}_2\cdot h/t_1+a\cdot t_2=h \\ h=\frac{a\cdot t_1\cdot t_2}{t_1-t_2}=\frac{10\cdot 0.5774\cdot 0.2679}{0.5774-0.2679}=5=5\text{m} \\ \text{ Verifying Solution: } \\ x=h/t_1=5/0.5774=5\sqrt{3}\text{m}\doteq 8.6603\text{m} \\ {\alpha }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x+a})=\frac{180°}{\pi }\cdot \arctan (\frac{5}{8.6603+10})=15{}^{\circ } \\ {\beta }_2=\frac{180°}{\pi }\cdot \arctan (\frac{h}{x})=\frac{180°}{\pi }\cdot \arctan (\frac{5}{8.6603})=30{}^{\circ } \end{gathered}$$

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