ANSWER
• BD = c * sin(A)
,
• BD = c * cos(B)
,
• BD = AD * tan(A)
,
• BD = AD/tan(B)
EXPLANATION
We have to use trigonometric ratios for triangle ABD, so we can ignore triangle BCD for now,
Side BD is opposite to angle A, and side AB is the hypotenuse of the triangle. If we write the sine of A,
[tex]\sin A=\frac{opposite}{hypotenuse}=\frac{BD}{AB}[/tex]
Solving for BD, and replacing AB by its length c,
[tex]BD=c\sin A[/tex]
Also, side BD is the adjacent side to the angle B, so we can also use the cosine of B,
[tex]\cos B=\frac{adjacent}{hypotenuse}=\frac{BD}{AB}[/tex]
Solving for BD,
[tex]BD=c\cos B[/tex]
Then, we could use the tangent of A,
[tex]\tan A=\frac{opposite}{adjacent}=\frac{BD}{AD}[/tex]
Solve for BD,
[tex]BD=AD\tan A[/tex]
Or the tangent of B,
[tex]\tan B=\frac{opposite}{adjacent}=\frac{AD}{BD}[/tex]
Solving for BD,
[tex]BD=\frac{AD}{\tan B}[/tex]