Question

The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are 30° and 45° respectively. Find the height of the tower and the distance between the tower and the building.

Answer 1

Let AB be the building and CD be the tower.

Draw AE ⊥ CD.



Suppose the height of the tower be h m.

Here, AB = 8 m

∠DAE = ∠ADX = 30º (Alternate angles)

∠DBC = ∠BDX = 45º (Alternate angles)

CE = AB = 8 m

∴ DE = CD − CE = (h − 8) m

Distance between the building and tower = BC

In right ∆BCD,

$$\begin{gathered} \tan 45°=\frac{\mathrm{CD}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{\mathrm{BC}} \\ \Rightarrow \mathrm{BC}=h\mathrm{m} \end{gathered}$$

In right ∆AED,

$$\begin{gathered} \tan 30°=\frac{\mathrm{DE}}{\mathrm{AE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h-8}{h}\left(\mathrm{AE}=\mathrm{BC}\right) \\ \Rightarrow \sqrt{3}h-8\sqrt{3}=h \end{gathered}$$
$$\begin{gathered} \Rightarrow \sqrt{3}h-h=8\sqrt{3} \\ \Rightarrow \left(\sqrt{3}-1\right)h=8\sqrt{3} \\ \Rightarrow h=\frac{8\sqrt{3}}{\sqrt{3}-1} \\ \Rightarrow h=\frac{8\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\times \left(\sqrt{3}+1\right)} \end{gathered}$$
$$\begin{gathered} \Rightarrow h=\frac{8\sqrt{3}\left(\sqrt{3}+1\right)}{2}\left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ \Rightarrow h=4\sqrt{3}\left(\sqrt{3}+1\right) \\ \Rightarrow h=\left(12+4\sqrt{3}\right)\mathrm{m} \end{gathered}$$

∴ Height of the tower = $\left(12+4\sqrt{3}\right)\mathrm{m}$

Also,

Distance between the tower and building = BC = $\left(12+4\sqrt{3}\right)\mathrm{m}$ (BC = h m)

Thus, the the height of the tower is $\left(12+4\sqrt{3}\right)\mathrm{m}$ and the distance between the tower and the building is $\left(12+4\sqrt{3}\right)\mathrm{m}$.