i know side AC = $\displaystyle \sqrt{52}$
but thats as far as my brain will take me right now, thanks
Hello,
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.
$\displaystyle \implies \frac{\color{red}BD}{BC}=\frac{BC}{AB}$
Does it help you ?
Hello, silencecloak!
There is a theorem which covers this very situation,28. Find $\displaystyle BD.$
Code:C * *| * * | * 12 * | * * | * * | * A * * * * * * * * * B D x : - - - - 14 - - - - - :
. . but we don't need it.
Let $\displaystyle x \,=\,BD$
Since $\displaystyle \Delta CDB \sim \Delta ACB\!:\;\;\frac{x}{12}\:=\:\frac{12}{14} \quad\Rightarrow\quad\boxed{ x \:=\:\frac{72}{7}}$