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P                    Q

10       A

S       24         R

We can use the Pythagorean Theorem to   find PR  =

√ [ SR^2 + PS^2]   = √ [ 576 + 100]  = √676  =  26

And  (1/2) of this  = PA  = SA  =  13

So.....using the Law of Cosines

PS^2  = PA^2 + SA^2  - 2(PA * SA) cos PAS

10^2 = 13^2 + 13^2 - 2(13 * 13) cos PAS

100 - 2(13)^2

___________  =   cos PAS

  -2 (13^2)

2(13)^2  - 100

___________   =   cos PAS

   2(13)^2

338 - 100

________    =    cos PAS

  338 

238                                119

___    =   cos PAS =     ____

338                                169

169 is the answer? How can cos be greater than 1?

You can draw triangle $PAS$ with $DS$ perpendicular(the altitude) to $AP$. You know $PS = 10$. Let's name the midpoint of $PS$ the letter $F$. $PF = 5, PA = 13,$ so $AF = 12$. Now, cos $PAS = \frac{AD}{AS}$. We already know $AS = 13,$ so we just need to solve for $AD$. To find $AD$, we first need to solve for $DS$, which is one of the altitudes of triangle $PAS$. We know that the area of a triangle is $\frac{1}{2}bh$, so we can make an equation. $\frac{1}{2}\times10\times12=\frac{1}{2}\times13\times DS$. From this, we get $DS = \frac{120}{13}$. Now, we can use Pythagorean Theorum to find $AD$. $12^2-(\frac{120}{13})^2=AD^2$. After calculating this, we can get $AD = \frac{12\sqrt{69}}{13}$. Now, lets plug this into our fraction. We can get $\frac{\frac{12\sqrt{69}}{13}}{13}$ which simplifies to $\frac{12\sqrt{69}}{169}$.

- Daisy