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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
avatar+87655 
+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°

 

 

cool cool cool

CPhill  Sep 4, 2017

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