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**