# sides of a triangle

• May 16th 2008, 12:14 AM
silencecloak
sides of a triangle
http://img505.imageshack.us/img505/1485/ddcvdsu4.jpg

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

but thats as far as my brain will take me right now, thanks
• May 16th 2008, 12:28 AM
Moo
Hello,

Quote:

Originally Posted by silencecloak
http://img505.imageshack.us/img505/1485/ddcvdsu4.jpg

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

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

How much do you know about trigonometry ?

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

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

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

• May 16th 2008, 12:44 AM
silencecloak
Thank you i appreciate your help as always
• May 16th 2008, 07:17 AM
Soroban
Hello, silencecloak!

Quote:

28. Find $BD.$

Code:

C
*
*|  *
* |    *  12
*  |        *
*  |          *
*    |              *
A *  *  *  *  *  *  *  *  * B
D        x
: - - - - 14  - - - - - :

There is a theorem which covers this very situation,
. . but we don't need it.

Let $x \,=\,BD$

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