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?
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) )