The perpendicular from vertex P of the triangle PQR meets QR at the point S. A point T is located on PR such that QT=TR=RS=1.

What is the length of PR, given that Angle QPR=90^o?

Consider triangle QTR, QT=TR=1 which implies that 
<TQR=<TRQ (let it be a) =a.
<PQR=90-<PRQ = 90-a
Therefore <PQT=90-a-a=(90-2a)
Consider triangle PTQ, apply sine rule
QT/sin90 = PT/sin(90-2a)
1/1=PT/cos2a
PT=cos2a
PR-1=2*cos^2(a)-1
PR=2*cos^2(a)----(1)
Consider triangle PRS, <SPR=90-<PRS=90-a
So, PR sin(90-a)=SR
=> PRcosa=1, so, sub. cos(a)=1/PR in eq(1)
PR=2*(1/PR)^2
PR^3=2 => PR = 2^(1/3) (or) cube-root(2)

Edited on August 7, 2007, 7:05 am

Posted by Praneeth Yalavarthi on 2007-07-10 07:13:14