# Thread: sides of a triangle

1. ## sides of a triangle

i know side AC = $\displaystyle \sqrt{52}$

but thats as far as my brain will take me right now, thanks

2. Hello,

Originally Posted by silencecloak

i know side AC = $\displaystyle \sqrt{52}$

but thats as far as my brain will take me right now, thanks
How much do you know about trigonometry ?

$\displaystyle \cos B=\frac{BC}{AB} \ \begin{array}{c} \leftarrow \text{adjacent} \\ \leftarrow \text{hypotenuse} \end{array}$ , considering triangle ABC.

$\displaystyle \cos B=\frac{\color{red}BD}{BC} \ \begin{array}{c} \leftarrow \text{adjacent} \\ \leftarrow \text{hypotenuse} \end{array}$ , considering triangle BCD.

$\displaystyle \implies \frac{\color{red}BD}{BC}=\frac{BC}{AB}$

3. Thank you i appreciate your help as always

4. Hello, silencecloak!

28. Find $\displaystyle BD.$

Code:
            C
*
*|  *
* |     *  12
*  |        *
*   |           *
*    |              *
A *  *  *  *  *  *  *  *  * B
D        x
: - - - - 14  - - - - - :
There is a theorem which covers this very situation,
. . but we don't need it.

Let $\displaystyle x \,=\,BD$

Since $\displaystyle \Delta CDB \sim \Delta ACB\!:\;\;\frac{x}{12}\:=\:\frac{12}{14} \quad\Rightarrow\quad\boxed{ x \:=\:\frac{72}{7}}$