2 Related Rate prob

A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 60 cm wide at the top, and has height 40 cm. If the trough is being filled with water at the rate of 0.1 m^3/min how fast is the water level rising when the water is 20 cm deep?
When the water level is x m deep, the width of the trapezoid at the water surface is
(0.6 - 0.2)*x/0.4 + 0.2 = (x + 0.2)m
Volume of water V = (1/2)[(x + 0.2) + 0.2]x10m^3
= (5x^2 + 2x) m^3
dV/dt = (10x + 2) (dx/dt)
dV/dt = 0.1 m^3/min, when x = 0.2 m,
0.1 = (2 + 2) (dx/dt)
dx/dt = 0.1/4 = 0.25 m/min
Two sides of a triangle have lengths 12 m and 17 m. The angle between them is increasing at a rate of 2 degree/min. How fast is the length of the third side increasing when the angle between the sides of fixed length is 60 degrees?
Let the angle be A and the side be a.
By cosine rule, cosA = (12^2 + 17^2 - a^2)/(2 * 12 * 17)
408cosA = 433 - a^2
When A = 60, a = 15.13
-408sinA(dA/dt) = -2a(da/dt)
dA/dt = 2 degree / min
= 2 * π / 180 radian / min = π/90 radian / min
-408(0.866)(π/90) = -2(15.13)(da/dt)
da/dt = 0.4076 m/min

2009-10-23 12:50:32 補充：
Correction : dx/dt = 0.1/4 = 0.025 m/min

2009-10-23 14:51:31 補充：
See if this picture helps:
http://img203.imageshack.us/img203/5567/calculus.png

(0.6 - 0.2)*x/0.4 + 0.2 = (x + 0.2)m
I don't understand this part. Can you explain it?