Question

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60° and 45°. Find the distance between the cars. (Take $\sqrt{3}=1.732$ )

Answer 1

Let AB be the tower of height 120 m.
C and D be two cars which are at angle of depression 45º and 60º respectively when observed from the top of the tower.
In $△$ADB,
$$\begin{gathered} \tan 60°=\frac{\mathrm{AB}}{\mathrm{DB}} \\ \Rightarrow \sqrt{3}=\frac{120}{\mathrm{DB}} \\ \Rightarrow \mathrm{DB}=\frac{120}{\sqrt{3}}.....\left(1\right) \end{gathered}$$
In $△$ACB,
$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{120}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=120.....\left(2\right) \end{gathered}$$
Distance between the two cars will be
DB + BC = $\frac{120}{\sqrt{3}}+120$
$$\begin{gathered} \Rightarrow \mathrm{DB}+\mathrm{BC}=120\left(\frac{1}{\sqrt{3}}+1\right) \\ =120\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \\ =120\times \frac{2.732}{1.732}=189.28\mathrm{m} \end{gathered}$$