Question

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Q.1. The angle of elevation of the top of a tower from a point A (on the ground) is 30$°$. On walking 50 m towards the tower, the angle of elevation is found to be 60$°$. Calculate:
(i) the height of the tower (correct to one decimal place).
(ii) the distance of the tower from A.

Answer 1

Dear student,
$$\begin{gathered} i)Lethbetheheightofthebuilding. \\ Inright\Delta ABC, \\ \tan 60°=\frac{DB}{BC} \\ \Rightarrow \sqrt{3}=\frac{h}{x} \\ \Rightarrow x=\frac{h}{\sqrt{3}}--[1] \\ In∆ABD \\ \tan 30°=\frac{DB}{AB} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\frac{h}{\sqrt{3}}+50}--(from1) \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\sqrt{3}h}{h+50\sqrt{3}} \\ \Rightarrow h+50\sqrt{3}=3h \\ \Rightarrow 50\sqrt{3}=2h \\ \Rightarrow h=25\sqrt{3} \\ \Rightarrow h=25\times 1.732 \\ \Rightarrow h=43.3m \\ thustheheightoftoweris43.3m. \\ ii)Dis\tan ceoftowerfrompointA=50+x \\ =50+\frac{h}{\sqrt{3}} \\ =50+\frac{25\sqrt{3}}{\sqrt{3}} \\ =50+25 \\ =75m \end{gathered}$$
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