cos(x)=AD/3, sin(x)=BD/3, cos(y)=CD/7, sin(y)=BD/7.

cos^{2}(x)+sin^{2}(x) = AD^{2}/9 + BD^{2}/9 = 1.

cos^{2}(y)+sin^{2}(y) = CD^{2}/49 + BD^{2}/49 = 1.

BD^{2} = 9 - AD^{2} = 49 - CD^{2}.

Remember that AD + CD = 9, or CD = 9 - AD.

9 - AD^{2} = 49 - (9-AD)^{2}.

9 = 49 - 81 + 18AD.

AD = 41/18.

(41/18)^{2} + BD^{2} = 9.

BD = (9-(41/18)^{2})^{1/2} = 1.95236

Others have pointed out an alternate solution using Heron's Formula, which states the area of a triange is sqr(s*(s-a)*(s-b)*(s-c)), where s=(a+b+c)/2, and a, b, and c are the lengths of the three sides.

In this case a, b, and c are 3, 7, and 9.

s=(3+7+9)/2=9.5

area=sqr(9.5*(9.5-3)*(9.5-7)*(9.5-9)) = sqr(1235/16) = 8.785641695

The area must also equal (1/2)*9*h, where h is the height.

So sqr(1235/16) = (1/2)*9*h

h=sqr(1235/16)*(2/9) = 1.95236

The Wizard of Odds