BK ⊥ AD
So, Δ ABK is a right triangle.
Therefore,
[tex]BK^{2} = AB^{2} - AK^{2}[/tex]
[tex]= 6^{2} - 3^{2}[/tex]
= 36 - 9
= 27
[tex]Hence, BK = 3\sqrt{3}[/tex] units
Also,
cos m∠A = [tex]\frac{3}{6}[/tex]
[tex]= \frac{1}{2}[/tex]
m∠A = [tex]60^{0}[/tex]
