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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 $PQS$?

AdminMod2 Sep 4, 2017

#1**+3 **

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°

CPhill Sep 4, 2017