Find DB in the diagram below.
[asy] size(250); pair A,B,C,D; A=origin; B=(4,0); C=3dir(30); D=(2.3,0); draw(C--A--B--C--D); dot("$A$",A,SW); dot("$B$",B,SE); dot("$C$",C,N); dot("$D$",D,S); label("$2$",A--D,S); label("$30^\circ$",A,6dir(15)); label("$45^\circ$",B,4dir(158)); label("$60^\circ$",C,5dir(-75)); [/asy]
Identify Right Triangles: Since ∠ABC=45∘, triangle ABC is a 45-45-90 triangle. This means ∠BAC=45∘ and ∠ACB=90∘. Additionally, since ∠CAD=30∘ and ∠ACB=90∘, triangle ACD is a 30-60-90 triangle.
Find AC in Triangle ACD: Because triangle ACD is a 30-60-90 triangle, we know that the ratio of side lengths is constant (i.e., AC/AD = 3). We are given that AD = 2, so:
AC = 3⋅AD=3⋅2=23
Find AB in Triangle ABC: Since triangle ABC is a 45-45-90 triangle, we know that the ratio of side lengths is also constant (i.e., AB/AC = 1). We found AC in the previous step, so:
AB = AC = 2√3
Find BD: Now that we know AB, we can find BD by subtracting AD from AB:
BD = AB - AD = 2√3 - 2
Therefore, the length of BD is 2*sqrt(3) - 2.