Consider the parallelogram PQRS with PS parallel to QR and PQ parallel to SR. The bisector of the angle PQR intersects PS at T.
Determine PQ, given that TS = 5, QT = 6 and RT = 6.
Call the measure of angle PQT by a. Call the length of PQ=SR=x
Angle PTQ is also a which means PT is also x.
Angle STR is also a.
By applying the law of cosines on triangles PQT and STR we have:
x^2 = x^2 + 6^2  2*x*6*cos(a)
x^2 = 5^2 + 6^2  2*5*6*cos(a)
Multiply the top equation by 5 and the bottom by x and adding yields:
5x^2  x^3 = 5x^2 25x + 180  36x
x^3  61x + 180 = 0
This cubic has 3 real roots {9,4,5}
x cannot equal 9 of course, it also cannot equal 5 because that would make triangles STR and PTQ congruent which would make PSRQ a rectangle.
So x=4

Posted by Jer
on 20070501 12:13:28 