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In triangle $ABC,$ $\angle C = 90^\circ.$  A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$  If the radius of the semicircle is equal to $1$ and $BC = \sqrt{3}$, then find $AB$.

 Mar 10, 2024
 #1
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Triangle COB is similar to Triangle DOB  by HL

 

tan (angle  COB) =   1/sqrt(3) = sqrt (3) / 3   = tan angle DOB

 

tan angle DBC = tan angle ABC

 

tan (angle ABC)  = tan (angle COB + angle DOB)  = tan (2 angle COB) = AC / BC

 

tan (angle COB + angle DOB )  =   [ tan COB + tan DOB ] / [ 1 - tan COB*tan DOB] =

 

[ 2sqrt (3)/3 ] / [ 1 - (sqrt (3) / 3)^2 ] =  (2sqrt (3)/3) / [ 1 - 1/3]  = (2sqrt(3)/3) / (2/3)  =

 

sqrt 3  =  AC / BC

 

sqrt 3  = AC / sqrt 3

 

sqrt 3 * sqrt 3  = AC  =  3

 

cool cool cool

 Mar 10, 2024

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