Hello, batman121!

A vertial pole stands on the horizontal ground.

A man stands east of the pole and measures the elevation of the top as 45°.

He moves due south 40 meters and it measures 42°.

What is the height of the pole? His first sighting looks like this:

Code:

P *
| *
| *
h | *
| *
| 45° *
Q * - - - - - * A
h

The man is at $\displaystyle A.$ .The pole is: $\displaystyle PQ = h$

He sights the top of the pole and finds $\displaystyle \angle PAQ = 45^o$

He moves 40 m south.

*Looking down at the ground*, the diagram looks like this: Code:

h
Q * - - - - - * A
* |
* | 40
* |
* B

Using Pythagorus: .$\displaystyle QB \:=\:\sqrt{h^2+40^2}$

Consider the triangle $\displaystyle PQB$ Code:

P *
| *
| *
h | *
| *
| 42° *
Q * - - - - - - - - * B
√(h² + 40²)

We have: .$\displaystyle \tan42^o \:=\:\frac{h}{\sqrt{h^2 + 1600}} \quad\Rightarrow\quad h \;=\;\sqrt{h^2+1600}\!\cdot\!\tan42^o $

Square both sides: . $\displaystyle h^2 \;=\;(h^2+1600)\tan^2\!42^o \quad\Rightarrow\quad h^2\;=\;h^2\!\cdot\!\tan^2\!42^o + 1600\!\cdot\!\tan^2\!42^o $

. . $\displaystyle h^2-h^2\!\cdot\!\tan^2\!42^o \;=\;1600\!\cdot\!\tan^2\!42^o \quad\Rightarrow\quad h^2(1 - \tan^2\!42^o) \;=\;1600\!\cdot\!\tan^2\!42^o$

Hence: .$\displaystyle h^2 \;=\;\frac{1600\!\cdot\!\tan^2\!42^o}{1-\tan^2\!42^o} \;=\;6853.417787 $

Therefore: .$\displaystyle h \;=\;82.785375182 \;\approx\;\boxed{82.8\text{ m}}$