http://i952.photobucket.com/albums/a...g/P5250091.jpg

The diagram shows an isosceles triangle $\displaystyle ABC$ with fixed length $\displaystyle AB$ and $\displaystyle AC$ of $\displaystyle 10cm$ each and Angle $\displaystyle ACB = $Angle $\displaystyle ABC = \theta $radians.

$\displaystyle A$ is a variable point which is at a height $\displaystyle h cm$ directly above the point $\displaystyle O$ while $\displaystyle B$ and $\displaystyle C$ are variable points which move horizontally along the line $\displaystyle l$.

Given that $\displaystyle A$ descends vertically towards the point $\displaystyle O$ such that the area of triangle $\displaystyle ABC$ is decreasing at a constant rate of $\displaystyle 0.7cm^2/s$, determine, at the instant when $\displaystyle A$ is $\displaystyle 6cm$ above $\displaystyle O$, the rate of change of $\displaystyle \theta$

$\displaystyle A=\frac{1}{2}(20cos\theta)h$

$\displaystyle =10hcos\theta$

$\displaystyle \frac{dA}{d\theta}=-10hsin\theta $

$\displaystyle when h=6, sin\theta =\frac{6}{10}$

$\displaystyle \theta=0.6435 rad$

$\displaystyle \frac{dA}{d\theta}=-10(6)(sin0.6435)$

$\displaystyle =-36$

$\displaystyle \frac{d\theta}{dt}=(\frac{dA}{dt})(\frac{d\theta}{ dA})$

$\displaystyle =(-0.7)(\frac{1}{-36})$

$\displaystyle =\frac{7}{360}$

but the answer is -0.025

And another part of the question asks : Find the rate at which C is moving away from O. I could not start