+0  
 
0
42
3
avatar+146 

The diagonals of rectangle PQRS intersect at point A. If PS = 10 and RS = 24, then what is cos PAS?

Mathgenius  Nov 18, 2018
 #1
avatar+92808 
+1

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

 

 

cool cool cool

CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
 #2
avatar+146 
0

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

Mathgenius  Nov 20, 2018
 #3
avatar+352 
0

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

dierdurst  Nov 21, 2018

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