Question

A man of height $1.5m$ sees the top of a tree and the angle of elevation of the top at his eye is $60°$. Find the height of the tree if the distance of the man from the tree is $36m$.

Answer 1

The distance of the man from the tree is $36m$.

Now, looking onto the diagram drawn below:

We came to know that $BC=DE=36m$

and, the height of the man $=1.5m$

i.e., $BD=CE=1.5m$

Now, from the $△ABC$,

$$\begin{gathered} tan\theta =\frac{AC}{BC} \\ \Rightarrow tan60°=\frac{AC}{36} \\ \Rightarrow AC=\tan 60°\times 36 \\ \Rightarrow AC=\surd 3\times 36 \\ \Rightarrow AC=1.732\times 36(Since\surd 3=1.732) \\ \Rightarrow AC=62.352 \end{gathered}$$

So, the height of the tree $=AC+CE$

$$\begin{gathered} =62.352+1.5 \\ =\mathbf{63}\mathbf{.}\mathbf{852} \end{gathered}$$

Hence, the height of the tree is $\mathbf{63}\mathbf{.}\mathbf{852}\mathbf{}\mathit{m}$.