Subject: question on trig
Name: will
Question: (a squared) * cot theta div 2b units Hi Will,Was there a diadram that went with this problem? I ask because your statement is true in some situations but not in others. Suppose the diagram is as below. The volume of water in the tank is the area of the triangular face times the length of the talk, L. If the base is returned to horizontal then the volume is the area of the rectangular face, b*d, times the length, L. Since the volume of water is unchanged the area of the triangular face in the first diagram must be b*d. But the area of the traiangle is x*a/2 = a^{2}/2*cot theta and thus b*d = a^{2}/2*cot theta. Thus d = (a^{2}cot theta)/(2*b). Now suppose that a = b and the front face of the the original diagram looks like the diagram below, with cot theta > 2. In this case your expression d = (a^{2}cot theta)/(2*b) gives a depth of d = (a^{2}cot theta)/(2*a) > a, which is impossible. Cheers,Harley
