+279 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 $[\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]$
+129899 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
+8 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 - ∃
