In triangle PQR, the respective length of the sides PQ and PR are denoted by u and v while the length of the median PS is denoted by w. It is known that w is the geometric mean of u and v, and Angle QPR = 60^o.

Determine |cos(Angle PQR) - cos(Angle QRP)|, where |x| denotes the absolute value of x.

Hi!

Let note QR=s

1. Because w is median we can write from the median theorem

w^2=(2(u^2+v^2)-s^2))/4 = uv (because w^2=uv),

so i find that   s=sqrt(2)|u-v|  (1)

2.Cos theorem for u and v give us

u^2=v^2+s^2-2vscos(R)

v^2=u^2+s^2-2uscos(Q)

so i find

|cos(Q)-cos(R)| = ((v^2-u^2+s^2)/v-(u^2-v^2+s^2)/u)/(2s) (2)

3. Cos theorem for s give us

s^2=u^2+v^2-2uvcos(P) and because cos(P)=1/2

s^2=u^2+v^2-uv (3)

After replacing (3) in (2), i find

|cos(Q)-cos(R)| = 3|u-v|/(sqrt(2)*s) = 3/(2*sqrt(2)) = 1.0606 

 

 

Edited on December 21, 2007, 5:10 pm

Posted by Chesca Ciprian on 2007-12-21 15:15:19