In triangle $PQR$, we have $\angle P = 90^\circ$, $QR = 20$, and $\tan R = 4\sin R$. What is $PR$?

QR is the hypotenuse of this right triangle....and we have that.....

tan R  = 4 / sin R

sinR * tanR  = 4

sin^2R / cosR =  4

(1 - cos^2R) / cosR  = 4

1 - cos^2R = 4cosR

cos^2R + 4cosR - 1  = 0

Let x = cosR  ........ so....

x^2 + 4x - 1   = 0

Solving this for x  gives  

x = - √5 - 2       or   x  =   √5 - 2

However....since R  is acute......the second value will only be good for the cosine

So x =  cos R  =  √5 - 2

So

Cos R  =  PR / QR

√5 - 2  =  PR / 20

So     ....    PR   =  20 ( √5 - 2)