Two post are 120 m apart from each other and height of 1 is double the other. From the mid point of the line joining feet of an observer finds an angular elevation of their tops to be complementary. Find the height of the post?

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :

Here , Post AB =  h  m so Post CD  =  2 h m .

We know :  tan $\mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$ , So

In $∆$ ABM we get

$$\begin{gathered} \tan \mathrm{\theta } = \frac{\mathrm{AB}}{\mathrm{AM}} \\ \Rightarrow \tan \mathrm{\theta } = \frac{h}{60} --- (1\hspace{0.17em}) \end{gathered}$$

And in $∆$ CDM we get

$$\begin{gathered} \tan \left( 90° - \mathrm{\theta } \right)= \frac{\mathrm{CD}}{\mathrm{CM}} \\ \Rightarrow \mathrm{Cot} \mathrm{\theta } = \frac{2 h}{60} (\mathrm{We} \mathrm{know} \tan \left( 90° - \mathrm{\theta } \right) = \mathrm{Cot} \mathrm{\theta } ) \\ \Rightarrow \frac{1}{\tan \mathrm{\theta }} = \frac{2 h}{60} ( \mathrm{We} \mathrm{know} \mathrm{Cot} \mathrm{\theta } = \frac{1}{\tan \mathrm{\theta }} ) \\ \Rightarrow \frac{1}{{\displaystyle \frac{h}{60}}} = \frac{2 h}{60} ( \mathrm{From} \mathrm{equation} 1 ) \\ \Rightarrow \frac{60}{{\displaystyle h}} = \frac{2 h}{60} \\ \Rightarrow 2 h^2 = 3600 \\ \Rightarrow h^2 = 1800 \\ \Rightarrow h = \sqrt{1800} \\ \mathbf{\Rightarrow }\mathit{h}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{30}\sqrt{\mathbf{2}} \end{gathered}$$
Therefore,

Height of posts  = 30 $\sqrt{2}$  m and 60$\sqrt{2}$  m                                                    ( Ans )

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