Hello, r_maths!

Welcome aboard!

Code:

* C
* |
A * * |
| * * |
| * * | 5
4 | * * |
| * θ * |
B * - - - - * - - - - - - * D
6 E 3 :

Two telephone poles, AB and CD, are situated 9m apart.

They are held by wires that are fixed to the ground at E.

Calculate exact value of $\displaystyle \cos(\angle AEC).$

We want to find $\displaystyle \cos\theta$ which is in $\displaystyle \Delta AEC.$

Look at the upper portion of the diagram. Code:

* C
| 1
A *-----------------------+ F
| 9 |
| |
4 | | 4
| |
B * - - - - + - - - - - - * D
6 E 3 :

Draw $\displaystyle AC.$

We have a right triangle $\displaystyle CFA$ with $\displaystyle AF = 9,\:CF = 1$

. . Hence: .$\displaystyle AC \:=\:\sqrt{9^2 + 1^2} \:=\:\sqrt{82}$

Back to the original diagram . . .

We have $\displaystyle \Delta AEC$ with $\displaystyle AE = \sqrt{52},\:CE = \sqrt{34},\:AC = \sqrt{82}$

Using the Law of Cosines, we have:

. . $\displaystyle \cos\theta \:=\:\frac{AE^2 + CE^2 - AC^2}{2(AE)(CE)} \:=\:\frac{(\sqrt{52})^2 + (\sqrt{34})^2 - (\sqrt{82})^2}{2(\sqrt{52})(\sqrt{34})}$ $\displaystyle = \:\frac{52 + 34 - 82}{2(2\sqrt{13})(\sqrt{34})}$

Therefore: .$\displaystyle \cos\theta \:=\:\frac{4}{4(\sqrt{13})(\sqrt{34})} \:=\:\frac{1}{\sqrt{442}} $