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

Nov 18, 2018

#1
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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   Nov 18, 2018
edited by CPhill  Nov 18, 2018
edited by CPhill  Nov 18, 2018
#2
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169 is the answer? How can cos be greater than 1?

Nov 20, 2018
#3
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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

Nov 21, 2018