Question

The angle of elevation of the top of a tower from a point on the ground, which is $64m$ away from the foot of the tower is $45°$. The height of the tower is

Answer 1

Let the height of the tower be $h$

$$\begin{gathered} In△ABC \\ \Rightarrow \tan \left(45°\right)=\frac{Oppositeside}{adjacentside} \\ \Rightarrow \tan \left(45°\right)=\frac{AC}{BC} \\ \Rightarrow 1=\frac{AC}{BC}\because \tan \left(45°\right)=1 \\ \Rightarrow AC=BC \\ \Rightarrow AC=64m \end{gathered}$$

Hence the height of the tower $AC=h=64m$