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1. The diagonals of rectangle PQRS intersect at point X. If PS = 10 and RS=24, then what is \(\cos\angle PXS \)

 

2. In a triangle ABC, take point D on \(\overline{BC}\) such that \(DB = 14, DA = 13, DC = 4,\)and the circumcircles of triangles ADB and ADC have the same radius. Find the area of triangle ABC 

 

3. Let $\mathcal{R}$ denote the circular region bounded by $x^2 + y^2 = 36$. The lines $x = 4$ and $y = 3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_1$, $\mathcal{R}_2$, $\mathcal{R}_3$, and $\mathcal{R}_4$. Let $[\mathcal{R}_i]$ denote the area of region $\mathcal{R}_i$. If $[\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4]$, then compute $[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$ 

 Apr 11, 2019
edited by yasbib555  Apr 11, 2019
 #1
avatar+100424 
+1

1.  See the image

 

X  = (12, 5)

 

Length of SX  = sqrt (12^2 + 5^2 ) = sqrt (169) = 13

Length of PX  = sqrt (12^2 + (10 - 5)^2 )  = sqrt (12^2 + 5^2) = sqrt (169) = 13

 

Using the Law of Cosines

 

RS^2 = PX^2 + SX^2 - 2(PX)(SX)cos(PXS)

 

10^2 = 13^2 + 13^2 - 2(13)(13) cos (PXS)

 

100 = 2(169) - 2(169)cos(PXS)

 

100 - 338

_________  = cos(PXS)

     -338

 

 

-238

____  = cos (PXS)

-338 

 

 

238                            119

___  = cos(PXS)  =   ____

338                            169

 

 

cool cool cool

 Apr 12, 2019
 #3
avatar+283 
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thank you so much!!

yasbib555  Apr 12, 2019
 #2
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Since No.1 has been solved by CPhill, I will solve No.2.

 

Solution No.2:

 

 

\(\angle ABC\) = \(\angle ACB\) which implies that \(ABC\) is isosceles. We drop the perpendicular \(M\) from \(A\) to \(BC\); we get that \(MC = 9\) and so \(MD = 5\) . Now by Pythagoras on \(AMD\)..., \(AM = 12\). Thus \(ABC\)'s areo equals to \(\frac{1}{2}(12*18 = 216) = \boxed{108}\) .

 

RB - ∃

 Apr 12, 2019
edited by RobertBoyle  Apr 12, 2019
 #4
avatar+283 
0

thank you so much too!!! 

yasbib555  Apr 12, 2019

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