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 cos AEC.

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You can't use a calculator....
I need help on how to arrive at the answer, without the aid of a calculator.
I know i will have to use something like:
"cos(alpha + beta)" = cos alpha x cos beta - sin alpha x sin beta

I know it sounds rude as it's my 1st post, but if your kind enough to help.

2. Originally Posted by r_maths
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 cos AEC.
--
You can't use a calculator....
I need help on how to arrive at the answer, without the aid of a calculator.
I know i will have to use something like:
"cos(alpha + beta)" = cos alpha x cos beta - sin alpha x sin beta

I know it sounds rude as it's my 1st post, but if your kind enough to help.
Hello,

I've modified your sketch a little bit. You can easily calculate the cosine values in the red and blue right triangles.

EB

3. thanks, il try now, if i need more help with this, please be on to help.

4. 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 $\cos(\angle AEC).$

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

Look at the upper portion of the diagram.
Code:
                              * C
| 1
A *-----------------------+ F
|           9           |
|                       |
4 |                       | 4
|                       |
B * - - - - + - - - - - - * D
6     E      3      :

Draw $AC.$
We have a right triangle $CFA$ with $AF = 9,\:CF = 1$
. . Hence: . $AC \:=\:\sqrt{9^2 + 1^2} \:=\:\sqrt{82}$

Back to the original diagram . . .

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

Using the Law of Cosines, we have:
. . $\cos\theta \:=\:\frac{AE^2 + CE^2 - AC^2}{2(AE)(CE)} \:=\:\frac{(\sqrt{52})^2 + (\sqrt{34})^2 - (\sqrt{82})^2}{2(\sqrt{52})(\sqrt{34})}$ $= \:\frac{52 + 34 - 82}{2(2\sqrt{13})(\sqrt{34})}$

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

5. Originally Posted by r_maths
thanks, il try now, if i need more help with this, please be on to help.
Hello,

I haven't got much time left. So I send you (only for confirmation of course) a possible way to do your problem:

From the drawing you can read:
$\cos(\alpha)=\frac{4}{\sqrt{52}}$
$\sin(\alpha)=\frac{6}{\sqrt{52}}$

$\cos(\beta)=\frac{5}{\sqrt{34}}$
$\sin(\beta)=\frac{3}{\sqrt{34}}$

Now use the formula you've mentioned above:

$\cos(\alpha+\beta)=\frac{4}{\sqrt{52}} \cdot \frac{5}{\sqrt{34}} - \frac{6}{\sqrt{52}} \cdot \frac{3}{\sqrt{34}}=$ $\frac{20}{\sqrt{52 \cdot 34}} - \frac{18}{\sqrt{52 \cdot 34}}=\frac{2}{\sqrt{52 \cdot 34}}=\frac{1}{\sqrt{442}}$

EB

6. ^
Soroban, thanks,
i was having difficulties with earboth solution. i got (19/442) or maybe its my bad maths skills.

thanks!

EDIT: thanks earboth

7. Originally Posted by r_maths
^
omg, thanks,
i was having difficulties with earboth solution. i got (19/442) or maybe its my bad maths skills.

thanks!
Hello,

you didn't subtract but you added the product of sine-values in your formula. That's all.

EB

8. yep, solved. thanks
im going to be a regular here if im going to past my exams