In the diagram, four circles of radius 1 with centres $P$, $Q$, $R$, and $S$ are tangent to one another and to the sides of $\triangle ABC$, as shown.
What is the degree measure of the smallest angle in triangle $PQS$?
Triangle PRS is equilateral.....so angle PSR = 60° = angle PSQ
And PS = 2 and QS = 4
And by the Law of Cosines we have that
PQ = sqrt ( 2*2 + 4^2 - 2(4)(2) cos 60 ) = sqrt ( 20 - 8) = sqrt (12) = 2sqrt(3)
And by the Law of Sines, we have
sin PSQ / PQ = sin PQS / PS
sin 60 / 2sqrt (3) = sin PQS / 2
(sqrt (3) / 2) / [ 2sqrt (3) ] = sin PQS / 2
(1/4) = sin PQS / 2
sin PQS = 1/2 ... thus PQS = 30°
So..since PS is the shortest side in triangle PQS, then angle PQS is the smallest angle = 30°