From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are…

In \(\displaystyle \triangle \mathrm{BPA}\) \[\begin{aligned} & \tan 45^{\circ}=1=\frac{20}{\mathrm{x}} \\ & \Rightarrow \mathrm{x}=20 \mathrm{~m} \end{aligned} \] Now, In △ CPA \[\begin{aligned} & \tan 60^{\circ}=\sqrt{3}=\frac{h+20}{\mathrm{x}} \\ & \Rightarrow \mathrm{~h}+20=\mathrm{x} \sqrt{3} \ldots \ldots \end{aligned} \] Solving (i) and (ii) \[h=20(\sqrt{3}-1) m \] ∴ Height of the tower is \(\displaystyle 20(\sqrt{3}-1) \mathrm{m}\).