+188 The diagonals of rectangle PQRS intersect at point x. if PS = 6 and RS = 8, then what is sin PXS?
+128408 P Q
6 X
S 8 R
This is similar to your other problem
Triangle PSR forms a 6-8-10 Pythagorean Right triangle with PR = 10
So....PX = SX = (1/2) (10) = 5
Using the Law of Cosines
PS^2 = PX^2 + SX^2 - 2(PX * SX)cosPXS
6^2 = 5^2 + 5^2 - 2 (5 * 5)cosPXS
36 - 2(5)^2
_________ = cos PXS
-2(5)^2
36 - 50
______ = cos PXS
-50
50 - 36 14 7
_____ = cos PXS = ___ = ___
50 50 25
Since PXS is acute.....then its sine will be positive and we can find it as
sin PXS = √ [ 1 - cos^2 PXS] = √ [ 1 - (7/25)^2 ] = √ [ 1 - 49/625] = √ [625 - 49]/ 25 =
√576/ 25 = 24 / 25
