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In triangle ABC, angle C = 90 degrees and angle CAB = 45 degrees.  Point D is inside the triangle such that angle CAD = 30 degrees, angle ADB = 150 degrees, and AD = 2.  What is BC?

 May 25, 2020
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Since  angle(CAB) = 45o  and  angle(CAD) = 30o,  angle(DAB) = 15o.

Since  angle(ADB) = 150o  and  angle(DAB) = 15o,  angle(DBA) = 15o.

 

Therefore,  AD = DB  and since  AD = 2,  DB = 2.

 

We can find the value of AB by using the Law of Cosines:

  AB2  =  AD2 + BD2 - 2·AD·BD·cos(ADB)   --->   AB2  =  22 + 22 - 2·2·2·cos(150o

    --->   AB2  =  4 + 4 - 8 · -sqrt(3)/2   --->   AB2  =  8 + 4sqrt(3)   

    --->   AB  =  sqrt( 8 + 4sqrt(3) )

 

Since triangle(ABC) is an isosceles right triangle,  BC  =  AB / sqrt(2)

                                                                                BC  =  sqrt( 8 + 4sqrt(3) ) / sqrt(2)

                                                                                BC  =  sqrt( 4 + 2sqrt(3) )

 May 25, 2020

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