Stuck on this trig question (1 Viewer)

[Kingportable]

http://i612.photobucket.com/albums/tt201/Ivanreyes/50b3b9f1629fe6b0b9e025b0cbd4c33c_zpsd7cfba57.jpg

Hey can please help me with this trig question.

An open rectangular tank a units deep and b units wide holds water and is tilted so that the base BC is returned to the horizontal. When BC is returned to the horizontal, show that the depth of the water is (a^2.cot(theta))/(2b) units.


I got up to b=a/costheta and went on to say...
that the area of the water is a^2/2CosTheta

which if I can find a is theta/b somewhere would be awesome lol
Anyways it's question 9 on the link I sent

 

[Drongoski]

Let dashed line thru A meet BC at D

angle ADB = angle theta (alt angles on // lines)

BD/a = cot angle ADB = cot theta ==> BD = a cot theta

Let depth be h

Now area of triangle ABD = area of rectangle: h x b (note: if width of tank is 'w', then area x w = volume, w same for both)

i.e. (1/2) x (a cot theta) x a = h x b

.: h = a² x cot theta/(2 x b)

 

[Kingportable]

Drogonski you legend