# Thread: Further maths trig question

1. ## Further maths trig question

the question give states " a vertial pole stands horizontal to the ground, a man stands east of the pole and measures the elevation of the top as 45 deg, he moves due south 40 meters and meausers 42 deg what is the height of the pole" after me and my mate tryed to draw it several times on a CAD program we got different answer each time varring from 381m-420m ish. i belive i have to use sin and cos rule to work this out but am having trouble.

the attachment contains a badly drawn example of what i assume it would look like.

and anybody tell me what i have to do in order to work this out please.

2. 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 $A.$ .The pole is: $PQ = h$
He sights the top of the pole and finds $\angle PAQ = 45^o$

He moves 40 m south.
Looking down at the ground, the diagram looks like this:
Code:
            h
Q * - - - - - * A
*        |
*     | 40
*  |
* B
Using Pythagorus: . $QB \:=\:\sqrt{h^2+40^2}$

Consider the triangle $PQB$
Code:
    P *
|  *
|     *
h |        *
|           *
|           42° *
Q * - - - - - - - - * B
√(h² + 40²)
We have: . $\tan42^o \:=\:\frac{h}{\sqrt{h^2 + 1600}} \quad\Rightarrow\quad h \;=\;\sqrt{h^2+1600}\!\cdot\!\tan42^o$

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

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

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

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