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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